HDOJ 2120 Ice_cream's world I（并查集）

Ice_cream's world I

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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

并查集的应用，用来查找被分割的区域个数。
即当两个节点值相同时说明已经为了一个圈，否则不可能，此时区域个数加1.
#include<stdio.h>  #include<algorithm>  #include<iostream>  using namespace std;  int a[10001000],k;  int find(int x)  {      if( a[x]!= x )       a[x] = find(a[x]);         return a[x];      }  void mange(int x,int y)  {      int fx,fy;      fx=find(x);      fy=find(y);      if(fx!=fy)      a[fx]=fy;    else    k++;//节点值相同则存在环    }  int main()  {      int n,m;  while(~scanf("%d%d",&n,&m))       {     for(int i=0;i<=n;i++)      {a[i]=i;}       k=0; int K1,K2;          for(int i=1;i<=m;i++)          {              scanf("%d%d",&K1,&K2);            mange(K1,K2);          }                    printf("%d\n",k);              }  }