道路修建

题目背景：

bzoj2435

分析：暴力dfs

突如其来的傻题get·····直接dfs，每一个点对答案的贡献是，abs((n – size[cur])- size[cur]) * w（w为这个点与父亲的连边的权值），然后就没有然后了······

Source：

/*
	created by scarlyw
*/
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cctype>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <ctime>

const int MAXN = 1000000 + 10;

long long ans = 0;
int n, x, y, z;
int size[MAXN];

struct node {
	int to, w;
	node(int to = 0, int w = 0) : to(to), w(w) {}
} ;

std::vector<node> edge[MAXN];

inline void add_edge(int x, int y, int z) {
	edge[x].push_back(node(y, z)), edge[y].push_back(node(x, z));	
}

inline void read_in() {
	scanf("%d", &n);
	for (int i = 1; i < n; ++i) scanf("%d%d%d", &x, &y, &z), add_edge(x, y, z);
}

inline void dfs(int cur, int fa, int w) {
	size[cur] = 1;
	for (int p = 0; p < edge[cur].size(); ++p) {
		node *e = &edge[cur][p];
		if (e->to != fa) dfs(e->to, cur, e->w), size[cur] += size[e->to];
	}
	ans += (long long)abs(n - size[cur] - size[cur]) * (long long)w;
}

int main() {
	read_in();
	dfs(1, 0, 0);
	printf("%lld", ans);
	return 0;
}