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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): 848 Accepted Submission(s): 494
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
Author
Wiskey
Source
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题意：判断有几个环
并查集判断成环
判断父节点是否相同 如果不同就有环
#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
int r[1010];
int flag;
int find_x(int x)
{
while(x!=r[x])
x=r[x];
return x;
}
void fun(int x,int y)
{
x=find_x(x);
y=find_x(y);
if(x!=y){
r[y]=x;
}
else flag++;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
for(int i=0;i<=n;i++)
r[i]=i;
flag=0;
int a,b;
for(int i=0;i<m;i++){
scanf("%d%d",&a,&b);
fun(a,b);
}
printf("%d\n",flag);
}
return 0;
}
HDOJ Ice_cream's world I 2120【并查集判断成环】
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原文地址：http://blog.csdn.net/ydd97/article/details/47156193
