Ponds

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1288    Accepted Submission(s): 429

Problem Description

Betty
owns a lot of ponds, some of them are connected with other ponds by
pipes, and there will not be more than one pipe between two ponds. Each
pond has a value v.

Now
Betty wants to remove some ponds because she does not have enough
money. But each time when she removes a pond, she can only remove the
ponds which are connected with less than two ponds, or the pond will
explode.

Note that Betty should keep removing ponds until no more
ponds can be removed. After that, please help her calculate the sum of
the value for each connected component consisting of a odd number of
ponds

Input

The first line of input will contain a number T(1≤T≤30) which is the number of test cases.

For each test case, the first line contains two number separated by a blank. One is the number p(1≤p≤104) which represents the number of ponds she owns, and the other is the number m(1≤m≤105) which represents the number of pipes.

The next line contains p numbers v1,...,vp, where vi(1≤vi≤108) indicating the value of pond i.

Each of the last m lines contain two numbers a and b, which indicates that pond a and pond b are connected by a pipe.

Output

For
each test case, output the sum of the value of all connected components
consisting of odd number of ponds after removing all the ponds
connected with less than two pipes.

Sample Input

1
7 7
1 2 3 4 5 6 7
1 4
1 5
4 5
2 3
2 6
3 6
2 7

Sample Output

21

Source

2015 ACM/ICPC Asia Regional Changchun Online

本题首先要删掉度数小于2的点，这里需要注意一个问题就是如果删掉一个点后产生另一个度数为小于二的点仍然需要删除，直到没有可以删除的点为止；

这里用vecotr和queue实现起来更方便一些；

最后再求连通块点数的时候用dfs即可

#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=10005;
const int maxm=100005;
int val[maxn];
vector<int> g[maxm];
int ind[maxn];
bool vis[maxn];
void dfs(int u,long long &cnt,long long &temp){
     vis[u]=true;
     cnt++;
     temp+=val[u];
           for(int i=0;i<g[u].size();i++){
             int v=g[u][i];
             if(vis[v])
                 continue;

                 dfs(v,cnt,temp);

           }
}

int main(){
  int t;
  scanf("%d",&t);
  while(t--){
       int n,m;
       memset(val,0,sizeof(val));
       memset(ind,0,sizeof(ind));
       memset(vis,false,sizeof(vis));

       scanf("%d%d",&n,&m);
       for(int i=1;i<=n;i++){
           g[i].clear();
         scanf("%d",&val[i]);
       }
       int u,v;
       for(int i=1;i<=m;i++){
         scanf("%d%d",&u,&v);
         g[v].push_back(u);
         g[u].push_back(v);
         ind[u]++;
         ind[v]++;
       }
       queue<int>q;
      for(int i=1;i<=n;i++ ){
          if(ind[i]<2){
             q.push(i);

          }
      }
      while(!q.empty()){
           int u=q.front();
           q.pop();
           vis[u]=true;
           ind[u]=0;
           for(int i=0;i<g[u].size();i++){
                int v=g[u][i];
                ind[v]--;
                if(ind[v]<2&&!vis[v]){
                q.push(v);

                }
           }
      }
      long long ans=0,cnt=0,temp=0;//注意，这里必须用long long，long int会wa
      for(int i=1;i<=n;i++){
         if(vis[i]==true)

             continue;
            temp=0;
            cnt=0;
            dfs(i,cnt,temp);
                if(cnt%2==1)
                ans+=temp;
      }
      printf("%lld\n",ans);
  }
  return 0;
}