POJ 3345 Bribing FIPA（树形DP）

很好的树形DP题目，借鉴了http://www.cppblog.com/Yuan/archive/2010/04/28/113833.html
以后有待于研究下红黑树了

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <vector>

#include <map>

#include <string>

#include <sstream>

using namespace std;



#define min(a,b) (((a) < (b)) ? (a) : (b))

const int MAXN = 210;

const int INF = 1e9;



vector<int> tree[MAXN];

int dp[MAXN][MAXN];             // dp[u][m] 节点u选择m个国家所花费的最小代价

int cost[MAXN], node[MAXN];     // node[i] = 1 represent i is the node of the tree, 0 represent the root of the tree

int n, m;



int dfs(int u)

{

    int child = 1;



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

        dp[u][j] = INF;

    dp[u][0] = 0;



    for (int i = 0; i < tree[u].size(); ++i)

    {

        int v = tree[u][i];

        child += dfs(v);

        for (int j = n; j > 0; --j)

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

                dp[u][j] = min(dp[u][j], dp[u][j-k] + dp[v][k]);

    }



    dp[u][child] = min(dp[u][child], cost[u]);

    return child;

}



int main()

{

    char str[1000];

    while (gets(str) && str[0] != '#')

    {

        sscanf(str, "%d %d", &n, &m);

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

            tree[i].clear();



        map<string, int> mapidx;

        int id = 0;



        memset(cost, 0, sizeof(cost));

        memset(node, 0, sizeof(node));



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

        {

            scanf("%s", str);

            if (mapidx.find(str) == mapidx.end())

                mapidx[str] = ++id;



            int u = mapidx[str];

            scanf("%d", &cost[u]);



            gets(str);

            stringstream s(str);

            string name;

            

            while (s >> name)

            {

                if (mapidx.find(name) == mapidx.end())

                    mapidx[name] = ++id;

                tree[u].push_back(mapidx[name]);

                ++node[mapidx[name]];

            }

        }

        cost[0] = INF;

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

            if (!node[i])

                tree[0].push_back(i);   // tree[0] is the root



        dfs(0);

        int ans = INF;

        for (int j = m; j <= n; ++j)

            if (dp[0][j] < ans)

                ans = dp[0][j];



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

    }

    return 0;

}