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\n", ans);

}

int main() {

    //freopen("test.in", "r", stdin);

    read(), sov();

    return 0;

}