2120 Ice_cream's world I

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本文介绍如何使用并查集解决ICE_cream'sworld中的土地划分问题，通过构建连接塔楼的桥梁来形成互不相交的土地环，并计算最多可以奖励给多少勤奋的ACM者。

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Ice_cream's world I

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

Total Submission(s): 842 Accepted Submission(s): 491

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

Sample Output

题意：

每组数据的第一个数表示有多少房子（编号从0~n-1），第二个数表示有多少个桥，后边有相应的数字，表示哪两个编号的房子之间有桥，求有多少个相互连通的小环.....

分析：

典型的并查集的问题，不过处理的时候特殊处理，遇到有环的时候（两个元素的的父节点相同），就累加变量加上 1 ，到最后输出这个结果.....

#include<stdio.h>
#include<string.h>
int per[1005],kase;//kase 就是累加变量，记录环的个数
void init(int n)//初始化
{
    for(int i=0;i<n;++i)
    {
        per[i]=i;
    }
}
int find(int x)//查找
{
    int r=x;
    while(per[r]!=r)
    {
        r=per[r];
    }
    int i=x,j;
    while(i!=r)//压缩
    {
        j=per[i];per[i]=r;i=j;
    }
    return r;
}
void join(int x, int y)
{
    int fx=find(x),fy=find(y);
    if(fx!=fy)//合并
    {
        per[fx]=fy;
    }
    else//已经连通的那么就累加变量！
    {
        ++kase;
    }
}
int main()
{
    int n,m,i,a,b;
    while(~scanf("%d%d",&n,&m))
    {
        kase=0;
        init(n);
        while(m--)
        {
            scanf("%d%d",&a,&b);
            join(a,b);//每次输入都合并一次
        }
        printf("%d\n",kase);
    }
    return 0;
}