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