2015 ACM/ICPC Asia Regional Changchun Online（1002）

Ponds

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)


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

就是拓扑排序，入度为0的不管，入度为1的删除同时和他相连的顶点入度减一。然后把联通分量的顶点数为奇数的加起来就可以了。

[code]#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<cstdio>
#include<ctime>
using namespace std;

const int N = 11111;
int n, q;
int a
;
int b
;
vector<int>g
;
bool flag
;

void f()//拓扑排序
{
    int i, j, k;
    int flag = 0;//只有更改过，下一次才需要更改
    while(!flag)
    {
        flag = 1;
        for(j=1; j<=n; ++j) //遍历所有的结点
        {
            if(b[j] == 1)
            {
                b[j]--; //该顶点的入度为-1，防止该顶点被再此遍历到
                for(k = 0; k < g[j].size(); k++)
                {
                    if(b[ g[j][k] ] > 0)
                    {
                        b[ g[j][k] ]--;
                        flag = 0;
                    }
                }
            }
        }
    }
}

long long dfs(int v, int& tmp)//返回联通分量的值，tmp得到联通分量的顶点个数
{
    tmp++;
    flag[v] = 1;
    int i;
    long long ans = a[v];
    for(i = 0; i < g[v].size(); i++)
    {
        if(b[ g[v][i] ] > 1 && !flag[ g[v][i] ])
        {
            ans += dfs(g[v][i], tmp);
        }
    }
    return ans;
}

int main(void)
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int i, j;
        memset(flag, 0, sizeof(flag));
        memset(b, 0, sizeof(b));
        scanf("%d%d", &n, &q);
        for(i = 1; i <= n; i++)     {scanf("%d", &a[i]);g[i].clear();}
        while(q--)
        {
            int x, y;
            scanf("%d%d", &x, &y);
            g[x].push_back(y);
            g[y].push_back(x);
            b[x]++;
            b[y]++;
        }
        long long ans = 0;
        f();
        for(i = 1; i <= n; i++)
        {
            if(!flag[i] && b[i] > 1)
            {
                int tmp = 0;
                long long temp = dfs(i, tmp);
                if(tmp&1) ans += temp;
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}