UVA - 12186 Another Crisis （树形DP）

         思路：dp[i]表示让上司i签字至少需要多少工人签字。

      转移方程：将i的所有节点根据所需工人数量升序排序，设i需要k个下属签字，dp[i] = sum{dp[v]| 0 <= v <k}

AC代码：

#include<cstdio>#include<algorithm>#include<cstring>#include<utility>#include<string>#include<iostream>#include<map>#include<set>#include<vector>#include<queue>#include<stack>using namespace std;#define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> const int maxn = 1e5 + 5;vector<int>son[maxn];int t;int dfs(int u) {	int n = son[u].size();	if(!n) return 1; //工人	vector<int>cnt;	for(int i = 0; i < n; ++i) {		int v = son[u][i];		cnt.push_back(dfs(v));	}	sort(cnt.begin(), cnt.end());	int c = (n*t - 1) / 100 + 1;	int ans = 0;	for(int i = 0; i < c; ++i) ans += cnt[i];	return ans;}int main() {	int n;	while(scanf("%d%d", &n, &t) == 2 && n) {		for(int i = 0; i <= n; ++i) son[i].clear();		int PRe;		for(int i = 1; i <= n; ++i) {			scanf("%d", &pre);			son[pre].push_back(i);		}		printf("%d/n", dfs(0));	}	return 0;}如有不当之处欢迎指出！