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
给出n个瞭望塔和各个塔之间的围墙数m,也给出了m组a,b表示两者间有围墙连接,求哨塔和围墙把土地分割成了几个部分。
题解
用到了并查集,就是看看能形成几个环,形成环的个数就是答案
代码
Select Code
#include<stdio.h>
int f[1001];
int gef(int v)
{
if(f[v]==v)
return v;
else
f[v]=gef(f[v]);
return f[v];
}
int mer(int x,int y)
{
int t1,t2;
t1=gef(x);
t2=gef(y);
if(t1!=t2)
{
f[t2]=t1;
return 0;
}
return 1;
}
int main()
{
int a,b,c,d,i;
while(~scanf("%d%d",&a,&b))
{
int ans=0;
for(i=0; i<a; i++)
f[i]=i;
for(i=0; i<b; i++)
{
scanf("%d%d",&c,&d);
ans=ans+mer(c,d);
}
printf("%d\n",ans);
}
return 0;
}