Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2856    Accepted Submission(s): 1619

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

Ice_cream的世界我

问题描述

冰淇淋的世界是一个富有的国家，它有许多肥沃的土地。今天，这位冰淇淋女王想把土地奖励给勤奋的ACMers。于是便设置了一些瞭望塔，并在瞭望塔之间筑起了墙，以划分冰淇淋的世界。但是，女王最多可以授予多少个ACMers是一个大问题。一个被墙包围的土地只能给一个ACMer并且不能越过墙，如果你能帮助女王解决这个问题，你将得到一块土地。

输入

在这种情况下，前两个整数N, M (N<=1000, M<=10000)表示瞭望塔的数量和墙的数量。瞭望塔的编号从0到N-1。接下来的M行，每一行包含两个整数A, B平均值A和B之间有一个墙(A和B是不同的)。以文件结束结束。

输出

输出将被授予的ACMers的最大数量。

一个回答，一行。

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

解题思路：把对应的灯塔看作顶点，把之间的围墙看作无向边，当整个图存在环的时候，形成一个地方。

代码：
#include<stdio.h>
int pre[1005];//pre[x]=y;树中x的前驱节点为y
int findset(int x)//x属于那个集合，即求x的根结点
{
    if(pre[x]!=x)
        return pre[x]=findset(pre[x]);
    else
        return x;
}
int union_set(int a,int b)//合并集合x，y
{
    int x=findset(a);
    int y=findset(b);
    if(x==y)
        return 1;
    else
    {
        pre[x]=y;//顺序无所谓,2棵树合并为1棵，1棵树的根节点指向另一颗的根节点
        return 0;
    }
}
int main()
{
    int i,n,m;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
	for(i=0;i<=n-1;i++)
	    pre[i]=i;
	int sum=0;
        for(i=1;i<=m;i++)
	{
	    int a,b;
	    scanf("%d %d",&a,&b);
	    sum+=union_set(a,b);
	}
	printf("%d\n",sum);
    }
    return 0;
}