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:
在瞭望塔之间筑墙分隔冰激淋世界,被墙围绕的为一个小的世界,可颁发ACMer一个奖章,墙所围绕的为一个环。
思路:
用并查集求环数,直接看代码!
代码:
#include<stdio.h>
int f[10010],n,m;
void init()
{
int i;
for(i=0; i<=n; i++)
f[i]=i;
}
int getf(int v)
{
if(f[v]==v)
return v;
else
{
f[v]=getf(f[v]);
return f[v];
}
}
int merge(int v,int u)
{
int t1,t2;
t1=getf(v);
t2=getf(u);
if(t1!=t2)
{
f[t2]=t1;
return 0;
}
return 1;
}
int main()
{
int i,x,y,s;
while(~scanf("%d %d",&n,&m))
{
init();
s=0;
init();
for(i=1; i<=m; i++)
{
scanf("%d %d",&x,&y);
s=s+merge(x,y);
}
printf("%d\n",s);
}
return 0;
}