bzoj 2435

http://www.lydsy.com/JudgeOnline/problem.php?id=2435

noi 你为什么那么diao， 这种世纪水题刷一道少一道啊。。。 我原来还以为是两边的联通块大小按已经连接上的点来算，然后发现是按照最后的联通块来算的(' '    ) 直接每个点 abs(n - 2 * size[x]) * dis(边权）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int maxn = 1000100;

struct edge {
    int t, d; 
    edge* next;
}e[maxn * 2], *head[maxn]; int ne = 0;
int n, m;

void addedge(int f, int t, int d) {
    e[ne].t = t, e[ne].d = d, e[ne].next = head[f], head[f] = e + ne ++;  
}

int sta[maxn], top = 0, size[maxn], dis[maxn];

void dfs(int x, int fa) {
    sta[++ top] = x; size[x] = 1;
    for(edge* p = head[x]; p; p = p-> next) {
        if(p-> t != fa) dis[p-> t] = p-> d, dfs(p-> t, x), size[x] += size[p-> t];
    }
}

int int_get() {
    int x = 0; char c = (char)getchar(); bool f =0 ;
    while(!isdigit(c)) {
        if(c == '-') f = 1;
        c = (char)getchar();
    }
    while(isdigit(c)) {
        x = x * 10 + (int)(c - '0');
        c = (char)getchar();
    }
    if(f) x = -x;
    return x;
}

void read() {
    n = int_get();
    for(int i = 1; i < n; ++ i) {
        int u, v, w; 
        u = int_get(), v = int_get(), w = int_get();
        addedge(u, v, w), addedge(v, u, w);
    }
}

long long ans = 0;

void sov() {
    dfs(1, 0);
    for(int i = top; i >= 2; -- i) {
        ans += (long long)(abs(n - 2 * size[sta[i]])) * (long long)dis[sta[i]];
    }
    printf("%lld
", ans);
}

int main() {
    //freopen("test.in", "r", stdin);
    read(), sov();
    return 0;
}