Luogu P2986 [USACO10MAR]Great Cow Gathering G

题目描述

https://www.luogu.com.cn/problem/P2986

比较基础的up and down(大概

Solution

我们令 f[u]表示u的子树中的所有点到u的距离和,g[u]表示整棵树上所有点到u的距离和

我们dfs两遍即可

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#include <iostream>
#include <cstdio>
#include <cstring>
#define maxn 100010
#define INF 1000000000000000000ll
#define ll long long
using namespace std;

int n, a[maxn], sum;

struct Edge {
int to, next, w;
} e[maxn * 2]; int c1, head[maxn];
inline void add_edge(int u, int v, int w) {
e[c1].to = v; e[c1].w = w;
e[c1].next = head[u]; head[u] = c1++;
}

int sz[maxn]; ll g[maxn], f[maxn];
void dfs(int u, int fa) {
sz[u] = a[u];
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w; if (v == fa) continue;
dfs(v, u); sz[u] += sz[v]; f[u] += f[v] + 1ll * sz[v] * w;
}
}

ll ans = INF;
void Dfs(int u, int fa) {
ans = min(ans, g[u]);
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w; if (v == fa) continue;
ll res = g[u] - 1ll * w * sz[v] - f[v] + 1ll * (sum - sz[v]) * w;
g[v] = res + f[v]; Dfs(v, u);
}
}

int main() { memset(head, -1, sizeof head);
scanf("%d", &n);
for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), sum += a[i];
for (int i = 1; i < n; ++i) {
int x, y, z; scanf("%d%d%d", &x, &y, &z);
add_edge(x, y, z); add_edge(y, x, z);
} dfs(1, 0); g[1] = f[1]; Dfs(1, 0);
cout << ans << endl;
return 0;
}