HDOJ 2120 Ice_cream's world I

Ice_cream’s world I

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

题意：构成了几个环。

题解：

void mix(int a,int b)
{
int x;
int y;
x=find(a);
y=find(b);
if(x!=y)
road[x]=y;
else
ans++;

}

这里这样写就OK

#include<stdio.h>
int road[1010];
int ans;
int find(int a)
{
if(road[a]==a) return a;
else
return find(road[a]);
}
void mix(int a,int b)
{
int x;
int y;
x=find(a);
y=find(b);
if(x!=y)
road[x]=y;
else
ans++;

}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
ans=0;
for(int i=0;i<=n-1;i++)
road[i]=i;
int a, b;
for(int i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
mix(a,b);
}
4000

printf("%d\n",ans);
}
}