pku 1639 Picnic Planning 最小度限制生成树

http://poj.org/problem?id=1639

题意：

见黑书。

思路：

最小限制树模板题：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <set>

#include <map>

#include <string>



#define CL(a,num) memset((a),(num),sizeof(a))

#define iabs(x)  ((x) > 0 ? (x) : -(x))

#define Min(a,b) (a) > (b)? (b):(a)

#define Max(a,b) (a) > (b)? (a):(b)



#define ll long long

#define inf 0x7f7f7f7f

#define MOD 1073741824

#define lc l,m,rt<<1

#define rc m + 1,r,rt<<1|1

#define pi acos(-1.0)

#define test puts("<------------------->")

#define maxn 100007

#define M 150

#define N 35

using namespace std;

//freopen("din.txt","r",stdin);



struct node

{

    int u,w;

    node(int _u,int _w)

    {

        u = _u;

        w = _w;

    }

    friend bool operator < (node a,node b)

    {

        return a.w > b.w;

    }

};



int dis[N],pre[N];

int col[N],near[N];

int Msd[N];

int mat[N][N];



priority_queue<node>q;

map<string,int>mp;



int res,n;



int prim(int s,int id)

{

    int res = 0,i;

    while (!q.empty()) q.pop();

    q.push(node(s,0));

    dis[s] = 0;

    while (!q.empty())

    {

        node cur = q.top(); q.pop();

        int u = cur.u;

        int w = cur.w;

        if (!col[u])

        {

            col[u] = id;

            res += w;

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

            {

                if (!col[i] && mat[u][i] && dis[i] > mat[u][i])

                {

                    dis[i] = mat[u][i];

                   pre[i] = u;

                    q.push(node(i,dis[i]));

                }

            }

        }

    }

    return res;

}

void dfs(int u,int fa,int maxside)

{

    Msd[u] = maxside > mat[fa][u] ? maxside : mat[fa][u];

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

    {

        if (mat[u][i] && i != fa && (pre[u] == i || pre[i] == u))

        dfs(i,u,Msd[u]);

    }

}



void solve(int k)

{

    int i;

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

    {

        dis[i] = inf;

        col[i] = 0;

        pre[i] = 0;

        near[i] = 0;

    }

    col[0] = 1;//标记每一个最小生成树，先将根节点计入1

    res = 0;

    int num = 0;

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

    {

        if (!col[i]) res += prim(i,++num);//分别计算最小生成树

    }

    //计算每个最小生成树里距离s的最短距离

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

    {

        int mk = col[i];

        if (mat[0][i] && (near[mk] == 0 || mat[0][i] < mat[0][near[mk]]))

        near[mk] = i;

    }

    //连接每个最小生成树到根节点s的最短距离同时更新每个节点可能连接的最大边权值

    CL(Msd,0);

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

    {

        res += mat[0][near[i]];

        mat[0][near[i]] = mat[near[i]][0] = 0;//将该点消除

        dfs(near[i],0,0);//求出改最小生成树以0为根节点的树根到每个节点之间的最大边权值

    }

    k = k - num;//此时形成num度最小生成树



    while (k--)

    {

        int tmp = 0;//找出Msd[tmp] - mat[0][tmp]的最大值

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

        {

            if (mat[0][i] && (tmp == 0 || Msd[tmp] - mat[0][tmp] < Msd[i] - mat[0][i]))

            tmp = i;

        }

        if (Msd[tmp] <= mat[0][tmp]) break;//如果不能使度限制生成树继续减小跳出

        res -= (Msd[tmp] - mat[0][tmp]);

        mat[0][tmp] = mat[tmp][0] = 0;//将改变置为0，表示不能再枚举了，已经加入最小限度生成树

        //

        int p = 0;

        for (i = tmp; pre[i] != 0; i = pre[i])

        {

            if (p == 0 || mat[p][pre[p]] < mat[i][pre[i]]) p = i;

        }

        pre[p] = 0;

        dfs(tmp,0,0);

    }

    printf("Total miles driven: %d\n",res);

}

int main()

{

    int i,m;

    string s1,s2;

    int w;

    while (~scanf("%d",&m))

    {

        mp.clear();

        CL(mat,0);

        mp["Park"] = 0; n = 0;

        //printf("%d\n",mp.count("Park"));

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

        {

            cin>>s1>>s2>>w;

            if (!mp.count(s1)) mp[s1] = ++n;

            if (!mp.count(s2)) mp[s2] = ++n;

            int tx = mp[s1];

            int ty = mp[s2];

            //建图

            if (!mat[tx][ty] || mat[tx][ty] > w)

            mat[tx][ty] = mat[ty][tx] = w;

        }

        int k;

        cin>>k;

        solve(k);

    }

    return 0;

}