HDU 3367 Pseudoforest

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3367

Problem Description

In graph theory, a pseudoforest is an undirected graph in which every connected component has at most one cycle. The maximal pseudoforests of G are the pseudoforest subgraphs of G that are not contained within any larger pseudoforest of G. A pesudoforest is larger than another if and only if the total value of the edges is greater than another one’s.

 

Input

The input consists of multiple test cases. The first line of each test case contains two integers, n(0 < n <= 10000), m(0 <= m <= 100000), which are the number of the vertexes and the number of the edges. The next m lines, each line consists of three integers, u, v, c, which means there is an edge with value c (0 < c <= 10000) between u and v. You can assume that there are no loop and no multiple edges.
The last test case is followed by a line containing two zeros, which means the end of the input.

 

Output

Output the sum of the value of the edges of the maximum pesudoforest.

 

Sample Input

3 30 1 11 2 12 0 14 50 1 11 2 12 3 13 0 10 2 20 0

 

Sample Output

35

题意： 要求每个连通分量最多可以有一个环

最小生成树的思想做

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;  //题意： 要求每个连通分量最多可以有一个环#define N 100005   //一定要开大一点int n,m;int father[N],vis[N];struct stud{int a,b;int len;}f[N];int cmp(stud a,stud b){    return a.len>b.len;}int cha(int x){    if(x!=father[x])        father[x]=cha(father[x]);    return father[x];}int fdd(int x,int y){    int xx=cha(x);    int yy=cha(y);    if(xx==yy)    {        if(vis[xx])            return 0;        vis[xx]=1;        return 1;    }    if(vis[xx]&&vis[yy])       return 0;    if(vis[xx])  //这一步非常重要，因为要把已经成环的    {              //点作为根节点        father[yy]=xx;        return 1;    }    else    {        father[xx]=yy;        return 1;    }}int main(){    int i;    while(scanf("%d%d",&n,&m)!=EOF,n||m)    {        for(i=0;i<=n;i++)            {                father[i]=i;                vis[i]=0;            }        for(i=0;i<m;i++)            scanf("%d%d%d",&f[i].a,&f[i].b,&f[i].len);        sort(f,f+m,cmp);        int ans=0;        for(i=0;i<m;i++)            if(fdd(f[i].a,f[i].b))               ans+=f[i].len;        printf("%d\n",ans);    }    return 0;}