HDU-2120-Ice_cream's world I

Ice_cream’s world I

Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 59 Accepted Submission(s) : 39

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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

题目意思：

有8个塔 有10条线（10 组数据 每组数据有两个数，意思是这两个塔之间有墙连着 也就是有十条线），解决的问题就是这十条线把这个封闭的区域分成了几小块。每条线之间还有可能相接，那么你所要做的就是这十条线能构成几个环，就是所形成的区域数

#include<cstdio>
#include<cstring>
#define maxn 1000+10
int p[maxn],ans=0;
int find(int x)
{
    while(x!=p[x])
        x=p[x];
    return p[x];
}
int he(int a,int b)
{
    if(find(a)!=find(b))
       p[find(a)]=find(b);
    else
        ans++;  // 如果根节点相同那么就成环了，++
    return  ans;
}
int main()
{
            int i,x,y;
            while(~scanf("%d%d",&x,&y))
            {
                for(i=0;i<x;i++)
                   p[i]=i;
                int x1,y1;
                while(y--)
               {
                    scanf("%d%d",&x1,&y1);
                    he(x1,y1);
               }
               printf("%d\n",ans);
               ans=0;
            }
    return 0;
}