首页 > ACM题库 > HDU-杭电 > HDU 3367-Pseudoforest-并查集-[解题报告]HOJ
2014
03-16

HDU 3367-Pseudoforest-并查集-[解题报告]HOJ

Pseudoforest

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

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

样例输出:

3
5

题意:找到一个这样的图,在这个图中,最多有一个环。

使得所有的边的和最大。

 

贪心+并查集

首先把边排序,然后开始分类讨论。

对于边ab(含有两个端点ab)

如果a,b是属于两个不同的集合

      a b 是两个环中的点,则放弃ab

      a b 有一个是环,则把环当做另一个的祖先,之后在回溯祖先的时候,能找到该点是在某个环中。

Pseudoforest

/*
 找到一个图,使得每一个连通分量最多有一个环
 */
 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #include<algorithm>
 using namespace std;
 const int maxn = 10005;
 const int maxm = 100005;
 struct node{
     int u,v,val;
 }edge[ maxm ];
 int fa[ maxn ],circle[ maxn ];
 int find( int x ){
     if( fa[x]==x ) return x;
     fa[x] = find(fa[x]);
     return fa[x];
 }
 bool union_ab( int x,int y ){
     int fax = find(x);
     int fay = find(y);
     if( fax==fay ){
         if( circle[ fax ]==-1 ){
             circle[ fax ] = 1;
             return true;
         }//形成一个环
         return false;
         //已经是环
     }
     else{
         if( circle[ fax ]==circle[ fay ]&&circle[ fax ]==1 )
             return false;
         if( circle[ fax ]==1 )
             fa[ fay ] = fax;
         else
             fa[ fax ] = fay;
         //这里注意把环作为祖先,因为find
         return true;
     }
 }
 void init( int n ){
     for( int i=0;i<n;i++ ){
         fa[i] = i;
         circle[ i ] = -1;
     }
 }
 int cmp( node a,node b ){
     return a.val>b.val;
 }
 int main(){
     int n,m;
     while( scanf("%d%d",&n,&m)==2,n||m ){
         //if( n==0&&m==0 ) break;
         for( int i=0;i<m;i++ )
             scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].val);
         init( n );
         sort( edge,edge+m,cmp );
         int ans = 0;
         for( int i=0;i<m;i++ ){
             if( union_ab( edge[i].u,edge[i].v) )
                 ans += edge[i].val;
         }
         printf("%d\n",ans);
     }
     return 0;
 }

View Code

 

参考:http://www.cnblogs.com/justforgl/archive/2013/07/07/3176517.html


  1. L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-1])这个地方也也有笔误
    应改为L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-2])