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有n个国家,你要获取m个国家的支持,获取第i个国家的支持就要给cost[i]的价钱 其中有一些国家是老大和小弟的关系,也就是说,如果你获得了某个老大国家的支持, 那么这个国家的所有小弟(包括小弟的小弟...递归下去)都会无偿免费支持你。 问最少的花费可以得到m个国家的支持
/**===================================================== * This is a solution for ACM/ICPC problem * * @source : poj-3345 Bribing FIPA * @description : 树形背包dp * @author : shuangde * @blog : blog.csdn.net/shuangde800 * @email : zengshuangde@gmail.com * Copyright (C) 2013/08/24 11:30 All rights reserved. *======================================================*/#include #include #include #include #include #include #include #include #include #include #define MP make_pair using namespace std; typedef pair PII; typedef long long int64; const int INF = 0x3f3f3f3f; const int MAXN = 210; vector adj[MAXN]; map name; int cost[MAXN]; int f[MAXN][MAXN]; int tot[MAXN]; bool deg[MAXN]; vector str[MAXN]; char buf[MAXN], buf2[MAXN]; int n, m; inline void read() { // clear for (int i = 0; i <= n; ++i) adj[i].clear(), str[i].clear(); name.clear(); int idx = 1; // read string buf; for (int i = 1; i <= n; ++i) { getline(cin, buf); stringstream sin(buf); string s; int v; sin >> s >> cost[i]; name[s] = i; while (sin >> s) { str[i].push_back(s); } } memset(deg, 0, sizeof(deg)); for (int i = 1; i <= n; ++i) { for (int e = 0; e < str[i].size(); ++e) { int v = name[str[i][e]]; adj[i].push_back(v); deg[v] = true; } } for (int i = 1; i <= n; ++i) if (!deg[i]) adj[0].push_back(i); } int dfs(int u) { tot[u] = 1; // count vertex for (int e = 0; e < adj[u].size(); ++e) { int v = adj[u][e]; tot[u] += dfs(v); } // init f[u][0] = 0; f[u][1] = cost[u]; if (u) { for (int i = 1; i <= tot[u]; ++i) f[u][i] = cost[u]; } // dp for (int e = 0; e < adj[u].size(); ++e) { int v = adj[u][e]; for (int i = tot[u]; i >= 1; --i) { for (int j = 1; j <= tot[v] && j <= i; ++j) f[u][i] = min(f[u][i], f[u][i-j] + f[v][j]); } } return tot[u]; } int main(){ while (gets(buf) && buf[0] != '#'){ sscanf(buf, "%d%d", &n, &m); read(); memset(f, INF, sizeof(f)); dfs(0); int ans = INF; for (int i = m+1; i <= n+1; ++i) ans = min(ans, f[0][i]); printf("%d\n", ans); } return 0; }