树形DP

切题ing!!!!!

HDU  2196 Anniversary party

经典树形DP,以前写的太搓了,终于学会简单写法了....

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <map>

#include <queue>

#include <set>

#include <vector>

#define MOD 100000000

#define LL __int64

using namespace std;

int dp[7000][2];

struct node

{

    int u,v,next;

}edge[7000];

int first[7000];

int flag[7000];

int p[7000];

int t;

void CL()

{

    t = 0;

    memset(first,-1,sizeof(first));

}

void add(int u,int v)

{

    edge[t].u = u;

    edge[t].v = v;

    edge[t].next = first[u];

    first[u] = t ++;

}

int dfs(int x,int st,int fa)

{

    int sum = 0,i,v;

    if(dp[x][st] != -1)

    return dp[x][st];

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        if(v == fa) continue;

        dfs(v,0,x);

        dfs(v,1,x);

        if(st == 0)

        sum += max(dp[v][0],dp[v][1]);

        else

        sum += dp[v][0];

    }

    if(st == 1)

    return dp[x][1] = sum + p[x];

    else

    return dp[x][0] = sum;

}

int main()

{

    int n,i,u,v;

    while(scanf("%d",&n)!=EOF)

    {

        CL();

        memset(dp,-1,sizeof(dp));

        for(i = 1;i <= n;i ++)

        {

            scanf("%d",&p[i]);

        }

        memset(flag,0,sizeof(flag));

        for(;;)

        {

            scanf("%d%d",&u,&v);

            if(!u&&!v) break;

            add(v,u);

            flag[u] = 1;

        }

        for(i = 1;i <= n;i ++)

        {

            if(!flag[i])

            add(0,i);

        }

        printf("%d\n",dfs(0,0,-1));

    }

    return 0;

}
View Code

HDU 2196 Computer

经典两次DFS,树形DP,也可以用最长路来做,我以前会的,忘了好多...

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <map>

#include <queue>

#include <set>

#include <vector>

#define MOD 100000000

#define LL __int64

using namespace std;

int dp[10001][2];

int ans[10001];

int flag[10001];

struct node

{

    int u,v,w,next;

}edge[20001];

int t;

int first[10001];

void CL()

{

    t = 1;

    memset(first,-1,sizeof(first));

}

void add(int u,int v,int w)

{

    edge[t].u = u;

    edge[t].v = v;

    edge[t].w = w;

    edge[t].next = first[u];

    first[u] = t ++;

}

void dfs1(int x)

{

    int i,v,max1 = 0,max2 = 0;

    flag[x] = 1;

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        if(flag[v]) continue;

        dfs1(v);

        if(max1 < dp[v][0] + edge[i].w)

        {

            max2 = max1;

            max1 = dp[v][0] + edge[i].w;

        }

        else if(max1 == dp[v][0] + edge[i].w)

        max2 = dp[v][0] + edge[i].w;

        else if(max2 < dp[v][0] + edge[i].w)

        max2 = dp[v][0] + edge[i].w;

    }

    dp[x][0] = max1;

    dp[x][1] = max2;

}

void dfs2(int x,int fa)

{

    int i,v;

    flag[x] = 1;

    ans[x] = max(dp[x][0],fa);

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        if(flag[v]) continue;

        if(dp[x][0] == dp[v][0] + edge[i].w)

        dfs2(v,max(fa,dp[x][1])+edge[i].w);

        else

        dfs2(v,max(fa,dp[x][0])+edge[i].w);

    }

}

int main()

{

    int i,n,v,w;

    while(scanf("%d",&n)!=EOF)

    {

        CL();

        for(i = 2;i <= n;i ++)

        {

            scanf("%d%d",&v,&w);

            add(v,i,w);

            add(i,v,w);

        }

        memset(flag,0,sizeof(flag));

        dfs1(1);

        //for(i = 1;i <= n;i ++)

        //printf("%d %d\n",dp[i][0],dp[i][1]);

        memset(flag,0,sizeof(flag));

        dfs2(1,0);

        for(i = 1;i <= n;i ++)

        printf("%d\n",ans[i]);

    }

    return 0;

}
View Code

HDU 1054 Strategic Game

以前做的时候,看错题了...这题是把所有的路,放一个士兵看守,那么如果根不放士兵,那么所有的子节点都要放。

#include <cstdio>

#include <cstring>

#include <string>

#include <iostream>

using namespace std;

#define INF 100000000

struct node

{

    int u,v,next;

}edge[1501];

int dp[1501][2];

int flag[1501];

int t;

int first[1501];

void CL()

{

    t = 0;

    memset(flag,0,sizeof(flag));

    memset(dp,0,sizeof(dp));

    memset(first,-1,sizeof(first));

}

void add(int u,int v)

{

    edge[t].u = u;

    edge[t].v = v;

    edge[t].next = first[u];

    first[u] = t ++;

}

void dfs(int x)

{

    int i,v;

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        dfs(v);

    }

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        dp[x][1] += min(dp[v][0],dp[v][1]);

        dp[x][0] += dp[v][1];

    }

    dp[x][1] ++;

}

int main()

{

    int i,j,n,m,u,v;

    while(scanf("%d",&n)!=EOF)

    {

        CL();

        for(i = 1;i <= n;i ++)

        {

            scanf("%d:(%d)",&u,&m);

            for(j = 1;j <= m;j ++)

            {

                scanf("%d",&v);

                flag[v] = 1;

                add(u,v);

            }

        }

        for(i = 0;i < n;i ++)

        {

            if(!flag[i])

            {

                dfs(i);

                printf("%d\n",min(dp[i][0],dp[i][1]));

                break;

            }

        }

    }

    return 0;

}
View Code

POJ 3107 Godfather

水题....

#include <cstdio>

#include <cstring>

#include <string>

#include <iostream>

using namespace std;

#define INF 100000000

struct node

{

    int u,v,next;

} edge[120001];

int first[51000];

int dp[51000];

int dis[51000];

int t;

void CL()

{

    t = 1;

    memset(first,-1,sizeof(first));

    memset(dp,0,sizeof(dp));

}

void add(int u,int v)

{

    edge[t].u = u;

    edge[t].v = v;

    edge[t].next = first[u];

    first[u] = t ++;

}

int n;

void dfs(int x,int fa)

{

    int i,v,temp = 1;

    for(i = first[x]; i != -1; i = edge[i].next)

    {

        v = edge[i].v;

        if(v == fa) continue;

        dfs(v,x);

        temp += dp[v];

    }

    dp[x] = temp;

    dis[x] = n - temp;

    for(i = first[x]; i != -1; i = edge[i].next)

    {

        v = edge[i].v;

        if(v == fa) continue;

        dis[x] = max(dis[x],dp[v]);

    }

}



int main()

{

    int i,u,v;

    while(scanf("%d",&n)!=EOF)

    {

        CL();

        for(i = 0; i < n-1; i ++)

        {

            scanf("%d%d",&u,&v);

            add(u,v);

            add(v,u);

        }

        dfs(1,-1);

        int ans = INF;

        for(i = 1; i <= n; i ++)

        {

            ans = min(ans,dis[i]);

        }

        int flag = 0;

        for(i = 1; i <= n; i ++)

        {

            if(ans == dis[i])

            {

                if(flag)

                    printf(" ");

                flag = 1;

                printf("%d",i);

            }

        }

        printf("\n");

    }

    return 0;

}
View Code

POJ 1155 TELE

分组背包,1Y哦,注意背的顺序,调了小会。。

#include <cstdio>

#include <cstring>

#include <string>

#include <iostream>

using namespace std;

#define INF 100000000

struct node

{

    int u,v,w,next;

} edge[120001];

int dp[3001][3001];

int p[3001];

int lf[3001];

int first[3001];

int t,n,m;

void CL()

{

    t = 1;

    memset(first,-1,sizeof(first));

    memset(p,0,sizeof(p));

    memset(lf,0,sizeof(lf));

}

void add(int u,int v,int w)

{

    edge[t].u = u;

    edge[t].v = v;

    edge[t].w = w;

    edge[t].next = first[u];

    first[u] = t ++;

}

void dfs(int x)

{

    int i,j,k,v,w,sum = 0;

    if(first[x] == -1)

    {

        dp[x][1] = p[x];

        lf[x] = 1;

        return ;

    }

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        dfs(v);

        sum += lf[v];

    }

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        w = edge[i].w;

        for(j = sum;j >= 1;j --)

        {

            for(k = 1;k <= min(j,lf[v]);k ++)

            {

                if(dp[v][k] == -INF) continue;

                dp[x][j] = max(dp[x][j],dp[v][k] + dp[x][j-k] - w);

            }

        }

    }

    lf[x] = sum;

}

int main()

{

    int i,j,k,v,w;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        CL();

        for(i = 1;i <= n;i ++)

        {

            for(j = 1;j <= n;j ++)

            dp[i][j] = -INF;

        }

        for(i = 1;i <= n-m;i ++)

        {

            scanf("%d",&k);

            for(j = 1;j <= k;j ++)

            {

                scanf("%d%d",&v,&w);

                add(i,v,w);

            }

        }

        for(i = n-m+1;i <= n;i ++)

        {

            scanf("%d",&p[i]);

        }

        dfs(1);

        for(i = m;i >= 1;i --)

        {

            if(dp[1][i] >= 0)

            break;

        }

        printf("%d\n",i);

    }

    return 0;

}
View Code

HDU 1011 Starship Troopers

分组背包,这个题意坑啊。如果想往下走,必须把根节点的bug消灭。

#include <cstdio>

#include <cstring>

#include <string>

#include <iostream>

using namespace std;

#define INF 100000000

struct node

{

    int u,v,next;

} edge[100001];

int p[101];

int w[101];

int first[101];

int flag[101];

int dp[101][101];

int t,n,m;

void CL()

{

    t = 1;

    memset(first,-1,sizeof(first));

    memset(dp,0,sizeof(dp));

}

void add(int u,int v)

{

    edge[t].u = u;

    edge[t].v = v;

    edge[t].next = first[u];

    first[u] = t ++;

}

void dfs(int x)

{

    int i,j,k,v,pre;

    flag[x] = 1;

    pre = (p[x]+19)/20;

    for(i = pre;i <= m;i ++)

    dp[x][i] = w[x];

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        if(flag[v]) continue;

        dfs(v);

        for(j = m;j >= pre;j --)

        {

            for(k = 1;k + pre <= j;k ++)//保证j-k>=pre,保证能消灭bug

            {

                dp[x][j] = max(dp[x][j],dp[v][k] + dp[x][j-k]);

            }

        }

    }

}

int main()

{

    int i,u,v;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        if(n < 0) break;

        CL();

        memset(flag,0,sizeof(flag));

        for(i = 1;i <= n;i ++)

        {

            scanf("%d%d",&p[i],&w[i]);

        }

        for(i = 1;i < n;i ++)

        {

            scanf("%d%d",&u,&v);

            add(u,v);

            add(v,u);

        }

        if(m == 0)

        {

            printf("0\n");

            continue;

        }

        dfs(1);

        printf("%d\n",dp[1][m]);

    }

    return 0;

}
View Code

HDU 1561 The more, The Better

挺简单的,注意 顺序和范围。

#include <cstdio>

#include <cstring>

#include <string>

#include <iostream>

using namespace std;

#define INF 100000000

struct node

{

    int u,v,next;

}edge[1001];

int first[301];

int dp[301][301];

int p[301];

int t,m;

void CL()

{

    t = 1;

    memset(first,-1,sizeof(first));

    memset(dp,0,sizeof(dp));

}

void add(int u,int v)

{

    edge[t].u = u;

    edge[t].v = v;

    edge[t].next = first[u];

    first[u] = t ++;

}

void dfs(int x)

{

    int i,j,k,v;

    for(i = 1;i <= m;i ++)

    dp[x][i] = p[x];

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        dfs(v);

        for(j = m;j >= 1;j --)

        {

            for(k = 1;k < j;k ++)

            dp[x][j] = max(dp[x][j],dp[v][k] + dp[x][j-k]);

        }

    }

}

int main()

{

    int n,i;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        if(n == 0&&m == 0) break;

        CL();

        m ++;

        for(i = 1;i <= n;i ++)

        {

            int u,v;

            scanf("%d%d",&u,&v);

            p[i] = v;

            add(u,i);

        }

        dfs(0);

        printf("%d\n",dp[0][m]);

    }

    return 0;

}
View Code

POJ 2486 Apple Tree

三种转移,少了一种....

#include <cstdio>

#include <cstring>

#include <string>

#include <iostream>

#include <vector>

using namespace std;

#define LL long long

struct node

{

    int u,v,next;

}edge[1000];

int p[300];

int first[300];

int flag[300];

int dp[300][300][2];

int t,n,m;

void CL()

{

    t = 1;

    memset(flag,0,sizeof(flag));

    memset(first,-1,sizeof(first));

    memset(dp,0,sizeof(dp));

}

void add(int u,int v)

{

    edge[t].u = u;

    edge[t].v = v;

    edge[t].next = first[u];

    first[u] = t ++;

}

void dfs(int x)

{

    int i,v,j,k;

    flag[x] = 1;

    dp[x][0][0] = dp[x][0][1] = p[x];

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        if(flag[v]) continue;

        dfs(v);

        for(j = m;j >= 0;j --)

        {

            for(k = 0;k <= j;k ++)

            {

                if(j-1-k >= 0)

                dp[x][j][0] = max(dp[x][j][0],dp[x][j-1-k][1]+dp[v][k][0]);

                if(j-2-k >= 0)

                {

                    dp[x][j][0] = max(dp[x][j][0],dp[x][j-2-k][0] + dp[v][k][1]);

                    dp[x][j][1] = max(dp[x][j][1],dp[x][j-2-k][1]+dp[v][k][1]);

                }

            }

        }

    }

}

int main()

{

    int u,v,i;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        CL();

        for(i = 1;i <= n;i ++)

        scanf("%d",&p[i]);

        for(i = 1;i < n;i ++)

        {

            scanf("%d%d",&u,&v);

            add(u,v);

            add(v,u);

        }

        dfs(1);

        int ans = 0;

        for(i = 0;i <= m;i ++)

        {

            ans = max(ans,dp[1][i][0]);

            ans = max(ans,dp[1][i][1]);

        }

        printf("%d\n",ans);

    }

    return 0;

}
View Code

 HDU 3586 Information Disturbing

乱搞。

#include <iostream>

#include <cstring>

#include <cstdio>

#include <queue>

#include <cstdlib>

#include <algorithm>

#include <queue>

using namespace std;

#define INF 1000000000

struct node

{

    int u,v,w,next;

}edge[2001];

int first[1001];

int flag[1001];

int dp[1001][1001];

int t,maxz;

void CL()

{

    t = 1;

    memset(first,-1,sizeof(first));

}

void add(int u,int v,int w)

{

    edge[t].u = u;

    edge[t].v = v;

    edge[t].w = w;

    edge[t].next = first[u];

    first[u] = t ++;

}

void dfs(int x,int fa)

{

    int v,i,j,temp,sum,num = 0;

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        if(v == fa) continue;

        dfs(v,x);

        num ++;

    }

    if(num == 0) return ;

    for(i = 1;i <= maxz;i ++)

    {

        sum = 0;

        for(j = first[x];j != -1;j = edge[j].next)

        {

            v = edge[j].v;

            if(v == fa) continue;

            if(i >= edge[j].w)

            temp = min(edge[j].w,dp[v][i]);

            else

            temp = dp[v][i];

            if(temp == INF)

            {

                dp[x][i] = INF;

                break;

            }

            sum += temp;

        }

        if(j == -1)

        dp[x][i] = sum;

    }

}

int main()

{

    int n,m,i,j;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        if(n == 0&&m == 0) break;

        CL();

        maxz = 0;

        for(i = 1;i < n;i ++)

        {

            int u,v,w;

            scanf("%d%d%d",&u,&v,&w);

            maxz = max(maxz,w);

            add(u,v,w);

            add(v,u,w);

        }

        for(i = 1;i <= n;i ++)

        {

            for(j = 1;j <= maxz;j ++)

            dp[i][j] = INF;

        }

        dfs(1,-1);

        for(i = 1;i <= maxz;i ++)

        {

            if(dp[1][i] <= m) break;

        }

        if(i == maxz + 1)

        printf("-1\n");

        else

        printf("%d\n",i);

    }

    return 0;

}
View Code

 HDU 4276 The Ghost Blows Light

感觉和吃苹果那个有点像,想了一个4维DP,然后妥妥的TLE了,然后想了想发现1到n,是必须走的,分着处理出来,减去这条必须走的,继续分组背包,写的很麻烦dp[i][j],表示以i为根花费j的回到i得到最大价值。

#include <iostream>

#include <cstring>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <map>

using namespace std;

int dp[101][501];

int flag[101];

int first[101];

int pre[101];

int p[101];

int sum[101];

int t,n,m;

struct node

{

    int u,v,w,next;

}edge[201];

void CL()

{

    t = 1;

    memset(first,-1,sizeof(first));

    memset(flag,0,sizeof(flag));

    memset(sum,0,sizeof(sum));

    memset(dp,0,sizeof(dp));

}

void add(int u,int v,int w)

{

    edge[t].u = u;

    edge[t].v = v;

    edge[t].w = w;

    edge[t].next = first[u];

    first[u] = t ++;

}

void find(int x,int fa)

{

    int i,v;

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        if(v == fa) continue;

        pre[v] = x;

        sum[v] = sum[x] + edge[i].w;

        find(v,x);

    }

}

void dfs(int x)

{

    int i,v,w,j,k;

    flag[x] = 1;

    dp[x][0] = p[x];

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        w = edge[i].w;

        if(flag[v]) continue;

        if(v == pre[x]) continue;

        dfs(v);

        for(j = m;j >= 0;j --)

        {

            for(k = 0;k <= j;k ++)

            {

                if(j-k-2*w >= 0)

                dp[x][j] = max(dp[x][j],dp[x][j-k-2*w] + dp[v][k]);

            }

        }

    }

}

int main()

{

    int u,v,w,i,k;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        CL();

        for(i = 1;i < n;i ++)

        {

            scanf("%d%d%d",&u,&v,&w);

            add(u,v,w);

            add(v,u,w);

        }

        for(i = 1;i <= n;i ++)

        scanf("%d",&p[i]);

        find(1,-1);

        if(sum[n] > m)

        {

            printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");

            continue;

        }

        int x = n;

        m -= sum[n];

        while(1)

        {

            dfs(x);

            for(i = m;i >= 0;i --)

            {

                for(k = 0;k <= i;k ++)

                {

                    dp[0][i] = max(dp[0][i],dp[0][i-k]+dp[x][k]);

                }

            }

            if(x == 1) break;

            x = pre[x];

        }

        int maxz = 0;

        for(i = 0;i <= m;i ++)

        maxz = max(maxz,dp[0][i]);

        printf("%d\n",maxz);

    }

    return 0;

}

/*

3 0

1 2 2

1 3 0

1 2 3

*/
View Code

 URAL 1389. Roadworks

尚大婶 说这题很水。。。必须水过去,dp[i][0]表示i点子树的道路没有被封,dp[i][1]表示i点在子树已经有道路被封了。

dp[i][0] = sum(dp[v][1]);

dp[i][1] = sum(dp[v][1]) + max(dp[v][0]-dp[v][1]) + 1;

找路径扯淡一点,爆栈,加代码就水过了。

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <iostream>

#include <cstring>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <map>

using namespace std;

int dp[300001][2];

int first[300001];

int flag[300001];

int t;

struct node

{

    int u,v,next;

}edge[300001];

int qu[300001];

int qv[300001];

int o[300001];

void CL()

{

    t = 1;

    memset(o,0,sizeof(o));

    memset(first,-1,sizeof(first));

    memset(flag,0,sizeof(flag));

    memset(dp,0,sizeof(dp));

}

void add(int u,int v)

{

    edge[t].u = u;

    edge[t].v = v;

    edge[t].next = first[u];

    first[u] = t++;

}

void dfs(int x)

{

    int i,v,maxz = -10000000,tf;

    flag[x] = 1;

    tf = 1;

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        if(flag[v]) continue;

        tf = 0;

        dfs(v);

        dp[x][0] += dp[v][1];

        maxz = max(maxz,dp[v][0]-dp[v][1]);

    }

    if(tf)

    {

        dp[x][0] = dp[x][1] = 0;

        return ;

    }

    dp[x][1] = dp[x][0] + maxz + 1;

}

void find(int x,int y,int fa)

{

    int i,v,maxz = -1000000,tf = 1,z;

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        if(v == fa) continue;

        maxz = max(maxz,dp[v][0]-dp[v][1]);

        tf = 0;

    }

    if(tf) return ;

    z = 1;

    for(i = first[x];i != -1;i = edge[i].next)

    {

        v = edge[i].v;

        if(v == fa) continue;

        if(y == 0)

        find(v,1,x);

        else

        {

            if(maxz == dp[v][0]-dp[v][1]&&z)

            {

                find(v,0,x);

                o[(i+1)/2] = 1;

                z = 0;

            }

            else

            find(v,1,x);

        }

    }

}

int main()

{

    int n,m,i;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        CL();

        for(i = 1;i <= m;i ++)

        {

            int u,v;

            scanf("%d%d",&u,&v);

            qu[i] = u;

            qv[i] = v;

            add(u,v);

            add(v,u);

        }

        dfs(1);

        printf("%d\n",dp[1][1]);

        find(1,1,-1);

        for(i = 1;i <= m;i ++)

        {

            if(o[i])

            printf("%d %d\n",qu[i],qv[i]);

        }

    }

    return 0;

}
View Code

 

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