Ice_cream's world I（并查集之判断有几个圈）

电玩女神 2022-08-05 05:19 70阅读 0赞

# Ice\_cream's world I #

**Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 746    Accepted Submission(s): 433**

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

Author

Wiskey

Source

[HDU 2007-10 Programming Contest\_WarmUp][HDU 2007-10 Programming Contest_WarmUp]

并查集本身是集合的应用，但是由于每次只告诉你两个点之间有墙，只要当这两个点的父节点是同一个，并且再次有墙时，就把结果加1，因为这个时候肯定就会产生一个区域。

#include<cstdio>
    #include<iostream>
    using namespace std;
    int a[10010],ans;
    int Find(int x)
    {
        int r=x;
        while(r!=a[r])
            r=a[r];
    
       /* int i=x,j;
        while(a[i]!=r)
        {
            j=a[i];
            a[i]=r;
            i=j;
        }*/
        return r;
    }//这个就是找老大的函数，什么时候停止？就是当一个人的老大是他自己的时候！
    void mix(int x,int y)
    {
        int fx=Find(x),fy=Find(y);//首先要找到他们各自的老大
        if(fx==fy)//如果他们的老大不相等，要合并就必须其中的一个老大向另外一个老大称老大
        {
            ans++;
        }
        else
        {
            a[fy]=fx;
        }
    }
    int main()
    {
        int n,m,i,b,c;
        while(cin>>n>>m)
        {
            ans=0;
            for(i=0;i<=n-1;i++)
            {
                a[i]=i;
            }
            for(i=1;i<=m;i++)
            {
                cin>>b>>c;
                mix(b,c);
            }
            cout<<ans<<endl;
        }
    }

[HDU 2007-10 Programming Contest_WarmUp]: http://acm.hdu.edu.cn/search.php?field=problem&key=HDU+2007-10++Programming+Contest_WarmUp&source=1&searchmode=source