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Ice_cream‘s world I
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 814 Accepted Submission(s): 478
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
思路：求环的个数。输入数据有多组。
代码：
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<queue>
using namespace std;
int per[1100];
int n,m,f;
int init()
{
for(int i=0;i<n;i++)
{
per[i]=i;
}
}
int find(int x)
{
int r;
r=x;
while(r!=per[r])
{
r=per[r];
}
per[x]=r;
return r;
}
int join(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
per[fx]=fy;
else
f++;
}
int main()
{
int a,b,i;
while(scanf("%d%d",&n,&m)!=EOF)
{
f=0;
init();
for(i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
join(a,b);
}
printf("%d\n",f);
}
return 0;
}
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hdoj 2120 Ice_cream's world I
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原文地址：http://blog.csdn.net/longge33445/article/details/47132351
