fzu2038 Another Postman Problem

思路：状态应该被狗日了，，，sum没有请0等错误然后wa了9发






因为这个是树，而且没有森林，所以呢，只需要求出每条边两端的点数，然后经过这条边的次数就等于两边点数的乘积，此时因为是有方向的左到右，右到左，所以要double下。

还有就是福大是win环境，要用I64d输出LL。

// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
inline int Readint(){
char c = getchar();
while(!isdigit(c)) c = getchar();
int x = 0;
while(isdigit(c)){
x = x * 10 + c - '0';
c = getchar();
}
return x;
}
struct edge
{
int v,cost;
};
vector<edge> G[100010];
LL sum;
bool vis[100010];
int n,res;
LL dfs(int u){
vis[u]=true;
LL res=0;
for (int i=0;i<G[u].size();++i){
int v = G[u][i].v;
if (vis[v])continue;
LL ans=dfs(v);
sum+=ans*(n-ans)*G[u][i].cost*2;
res+=ans;
}
return res+1;
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int icase=0;
int t;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
int a,b,c;
sum=0LL;
memset(vis, false,sizeof vis);
for (int i=0;i<n;++i)G[i].clear();
for (int i=1;i<n;++i){
scanf("%d%d%d",&a,&b,&c);
edge x,y;
x.v=b,x.cost=c;
y.v=a,y.cost=c;
G[a].push_back(x);
G[b].push_back(y);
}
dfs(0);
printf("Case %d: ",++icase);
printf("%I64d\n",sum);
}
return 0;
}