HDOJ 題目3367 Pseudoforest（并查集）
來源：程序員人生 發布時間：2015-07-29 08:17:27 閱讀次數：3287次
Pseudoforest
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1932 Accepted Submission(s): 746
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 3
0 1 1
1 2 1
2 0 1
4 5
0 1 1
1 2 1
2 3 1
3 0 1
0 2 2
0 0
Sample Output
Source
“光庭杯”第5屆華中北區程序設計約請賽
暨 WHU第8屆程序設計比賽
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ac代碼
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int pre[10010],vis[10010];
struct s
{
int u,v,w;
}edge[100100];
int n,m;
int cmp(const void *a,const void *b)
{
return (*(struct s *)b).w-(*(struct s *)a).w;
}
void init()
{
int i;
for(i=0;i<=n;i++)
pre[i]=i;
}
int find(int x)
{
if(x==pre[x])
return x;
return pre[x]=find(pre[x]);
}
int merge(int u,int v)
{
int fu=find(u);
int fv=find(v);
if(fu==fv)
{
if(!vis[fu])
{
vis[fu]=1;
return 1;
}
return 0;
}
if(vis[fu]&&vis[fv])
return 0;
if(vis[fu])
pre[fv]=fu;
else
pre[fu]=fv;
return 1;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF,n||m)
{
int i;
init();
for(i=0;i<m;i++)
{
int u,v,w;
scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
}
qsort(edge,m,sizeof(edge[0]),cmp);
int ans=0;
memset(vis,0,sizeof(vis));
for(i=0;i<m;i++)
{
int u=edge[i].u;
int v=edge[i].v;
int w=edge[i].w;
if(merge(u,v))
ans+=w;
}
printf("%d
",ans);
}
}
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