FZU - 2038 Another Postman Problem

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题意：对于每一条边，它的访问次数等于它子树的节点个数*剩余的节点个数

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int MAXN = 100010;struct node{    int v,w;    node(){}    node(int a,int b){        v = a;        w = b;    }};int vis[MAXN],T,n;vector<node> G[MAXN];long long ans;long long cal(int u){    vis[u] = 1;    int size = G[u].size();    long long sum = 1,temp;    for (int i = 0; i < size; i++){        int v = G[u][i].v;        if (vis[v])            continue;        temp = cal(v);        ans += 2 * G[u][i].w * temp * (n-temp);        sum += temp;    }    return sum;}int main(){    int t,u,v,w,cas=1;    scanf("%d",&t);    while (t--){        for (int i = 0; i < MAXN; i++)            G[i].clear();        memset(vis,0,sizeof(vis));        scanf("%d",&n);        for (int i = 1; i < n; i++){            scanf("%d%d%d",&u,&v,&w);            G[u].push_back(node(v,w));            G[v].push_back(node(u,w));        }        ans = 0;        cal(0);        printf("Case %d: %lld\n",cas++,ans);    }    return 0;}

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