BZOJ2435——[Noi2011]道路修建

1、题意：给个树，边的权值=两边的点数差*此边的长度，求所有边的权值和

2、分析：真不想说啥了。。。dfs即可

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
#define M 2000010

inline int read(){
    char ch = getchar(); int x = 0, f = 1;
    while(ch < '0' || ch > '9'){
        if(ch == '-') f = -1;
        ch = getchar();
    }
    while('0' <= ch && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

struct Edge{
    int u, v, w, next;
} G[M];
int head[M], tot;
int size[M];
LL ans;
int n;

inline void add(int u, int v, int w){
    G[++ tot] = (Edge){u, v, w, head[u]};
    head[u] = tot;
}

inline void dfs(int x, int fa){
    size[x] = 1;
    for(int i = head[x]; i != -1; i = G[i].next) if(G[i].v != fa){
        dfs(G[i].v, x);
        size[x] += size[G[i].v];
        ans += (LL)abs(n - size[G[i].v] - size[G[i].v]) * (LL)G[i].w;
    }
}

int main(){
    n = read();
    memset(head, -1, sizeof(head));
    for(int i = 1; i < n; i ++){
        int u = read(), v = read(), w = read();
        add(u, v, w); add(v, u, w);
    }
    dfs(1, 0);
    printf("%lld\n", ans);
    return 0;
}