UVA - 12186 Another Crisis (树形DP)

时间:2021-11-25 07:29:02

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!