hdoj--2120--Ice_cream's world I（并查集判断环）

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 975    Accepted Submission(s): 567

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

8 100 11 21 32 43 40 55 66 73 64 7

 

Sample Output

3

 

Author

Wiskey

 

Source

HDU 2007-10 Programming Contest_WarmUp

 

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好久不写并查集，刚开始以为会是scc什麽的，还好没写- -||

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int pre[10010],n,m,cnt;void init(){for(int i=0;i<10010;i++)pre[i]=i;}int find(int x){return pre[x]==x?x:find(pre[x]);}void join(int x,int y){int fx=find(x);int fy=find(y);if(fy!=fx)pre[fy]=fx;elsecnt++;}int main(){while(cin>>n>>m){init();int a,b;cnt=0;for(int i=0;i<m;i++){cin>>a>>b;join(a,b);}cout<<cnt<<endl;}return 0;}