P2052 [NOI2011]道路修建
题目:
解析:
维护一下每个子树的\(size\),这条边的贡献就是\(((n-size[v])-size[v])\times w=(n-2\times size[v])\times w\),\(v\)是这条边所到达的点,\(w\)是边权
代码:
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 2e6 + 10;
int n, m, num, ans;
int head[N], size[N];
struct node {
int v, nx, w;
} e[N];
inline void add(int u, int v, int w) {
e[++num] = (node) {v, head[u], w}, head[u] = num;
}
void dfs(int u, int fa) {
size[u] = 1;
for (int i = head[u]; ~i; i = e[i].nx) {
int v = e[i].v;
if (v == fa) continue;
dfs(v, u);
ans += abs(n - 2 * size) * e[i].w;
size[u] += size[v];
}
}
signed main() {
ios::sync_with_stdio(false);
memset(head, -1, sizeof head);
cin >> n;
for (int i = 1, x, y, z; i < n; ++i)
cin >> x >> y >> z, add(x, y, z), add(y, x, z);
dfs(1, 0);
cout << ans;
}
如果哪里有错误或不易理解,还请不吝赐教