首页 > 代码库 > HDU 2120 Ice_cream's world I【并查集】
HDU 2120 Ice_cream's world I【并查集】
解题思路：给出n对点的关系，求构成多少个环，如果对于点x和点y,它们本身就有一堵墙，即为它们本身就相连，如果find(x)=find(y)，说明它们的根节点相同，它们之间肯定有直接或间接的相连，即形成环
样例的示意图
共3个环
Ice_cream‘s world I
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 642 Accepted Submission(s): 371
Problem Description
ice_cream‘s world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.
Input
In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.
Output
Output the maximum number of ACMers who will be awarded. One answer one line.
Sample Input
8 10 0 1 1 2 1 3 2 4 3 4 0 5 5 6 6 7 3 6 4 7
Sample Output
3
#include<stdio.h>#include<string.h> int pre[10005],a[10005]; int find( int root){ if(root!=pre[root]) pre[root]=find(pre[root]); return pre[root];}void unionroot( int root1,int root2){ int x,y; x=find(root1); y=find(root2); if(x!=y) pre[x]=y;}int main(){ int root1,root2,x,y,i,n,m,tmp; while(scanf("%d %d",&n,&m)!=EOF) { tmp=0; for(i=0;i<=10005;i++) pre[i]=i; while(m--) { scanf("%d %d",&root1,&root2); x=find(root1); y=find(root2); if(x==y) tmp++; unionroot(x,y); } printf("%d\n",tmp); }}
HDU 2120 Ice_cream's world I【并查集】
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