4732: 【GESP2512六级】路径覆盖

内存限制:512 MB 时间限制:1.000 S
评测方式:文本比较 命题人:
提交:1 解决:1

题目描述

样例输入 复制

4
1 2 3
5 6 2 3

样例输出 复制

2

提示

#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e5 + 5;
int n;
int f[N], c[N], cnt[N];
long long ans[N];
int main() {
    scanf("%d", &n);
    for (int i = 2; i <= n; i++) {
        scanf("%d", &f[i]);
        cnt[f[i]]++;
    }
    for (int i = 1; i <= n; i++)
        scanf("%d", &c[i]);
    for (int i = n; i >= 1; i--) {
        if (cnt[i] == 0)
            ans[i] = c[i];
        ans[i] = min(ans[i], 1ll * c[i]);
        ans[f[i]] += ans[i];
    }
    printf("%lld\n", ans[1]);
    return 0;
}