Ponds

Ponds

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1394    Accepted Submission(s): 467

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

题意：有很多池塘，有的两个池塘间有管道连着。经费问题，要删除一些就连着一个管道的池塘。注意删除一个池塘之后，原本可能不用删除的，可能需要删除。4连着两个池塘2和3,2被删除后，4也要删除，因为连着4的管道由之前的2变成1个了。问剩下的在一个联通块里边池塘个数是奇数的，池塘价值之和。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<queue>
 #include<stack>
 #include<vector>
 #include<algorithm>

 using namespace std;

 #define maxn 10007

 vector<vector<int> > G;
 int used[maxn];
 int a[maxn*], b[maxn*];
 int f[maxn], v[maxn], sum[maxn], r[maxn];
 long long ans;

 int found(int x)
 {
     if(f[x] != x)
         f[x] = found(f[x]);
     return f[x];
 }

 int main()
 {
     int c, n, m;
     scanf("%d", &c);
     queue<int> Q;

     while(c--)
     {
         ans = ;
         G.resize(maxn+);
         G.clear();
         memset(used, , sizeof(used));
         memset(sum, , sizeof(sum));
         memset(a, , sizeof(a));
         memset(b, , sizeof(b));

         scanf("%d%d", &n, &m);
         for(int i = ; i <= n; i++)
         {
             f[i] = i;
             r[i] = ;
             scanf("%d", &v[i]);
         }

         for(int i = ; i < m; i++)
         {
             scanf("%d%d", &a[i], &b[i]);
             r[a[i]]++, r[b[i]]++;
             G[a[i]].push_back(b[i]);
             G[b[i]].push_back(a[i]);
         }
         for(int i = ; i <= n; i++)
             if(r[i] < )
                 Q.push(i);

         while(Q.size())
         {
             int u = Q.front();
             Q.pop();
             used[u] = ;
             int len = G[u].size();
             for(int i = ; i < len; i++)
             {
                 int v = G[u][i];
                 r[v]--;
                 if(r[v] <  && !used[v])
                     Q.push(v);
             }
         }
         for(int i = ; i < m; i++)
         {
             if(!used[a[i]] && !used[b[i]])
             {
                 int u = found(a[i]);
                 int v = found(b[i]);
                 if(u != v)
                     f[v] = u;
             }
         }
         for(int i = ; i <= n; i++)
             if(!used[i])
                 sum[found(i)]++;

         for(int i = ; i <= n; i++)
         {
             if(!used[i] && sum[f[i]] % )
                 ans += v[i];
         }
         printf("%I64d\n", ans);
     }
     return ;
 }