HDU 5438（并查集 + dfs）

Ponds

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3544    Accepted Submission(s): 1055

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

 

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题意就是删除度数小于2的点，输出含有奇数个点的连通分量的权值。用并查集找连通分量，dfs删除度数小于二的点。

#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
map<int, int> c;
int a[10010];
int p[10010];
set<int> g[10010];
int find(int x){
    return x == p[x] ? x : find(p[x]);
}
void dfs(int u){
    int t = *g[u].begin();
    g[u].erase(t);
    g[t].erase(u);
    if(g[t].size() == 1) dfs(t);
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        c.clear();
        memset(p, 0, sizeof p);
        memset(a, 0, sizeof a);
        int n, m;
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        for(int i = 1; i <= n; i++) g[i].clear();
        for(int i = 1; i <= n; i++) p[i] = i;
        for(int i = 0; i < m; i++){
            int x, y;
            scanf("%d%d", &x, &y);
            g[x].insert(y);
            g[y].insert(x);
            int xx = find(x);
            int yy = find(y);
            if(xx != yy){
                p[xx] = yy;
            }
        }
        for(int i = 1; i <= n; i++)
            if(g[i].size() == 1) dfs(i); // 删除度数等于1的点
        for(int i = 1; i <= n; i++) {if(g[i].size() > 1)++c[find(i)];}
        ll ans = 0;
        for(int i = 1; i <= n; i++){
            if(c[find(i)] % 2 && g[i].size() > 1)
                ans += a[i];
        }
        printf("%lld\n", ans);
    }
}