并查集 HDU 2120

其实就是判断新加的点是否在同一个集合里面

代码虐我千百遍，我待代码入初恋

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 583    Accepted Submission(s): 333

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

8 100 11 21 32 43 40 55 66 73 64 7

 

Sample Output

3

 

#include <stdio.h>#include <stdlib.h>#include <malloc.h>#include <limits.h>#include <ctype.h>#include <string.h>#include <string>#include <queue>#include <algorithm>#include <iostream>#include <stack>#include <deque>#include <vector>#include <set>#include <map>using namespace std;#define MAXN 1000 + 10#define MAXN1 10000 + 10int father[MAXN];int a[MAXN1];int b[MAXN1];int find(int x){    if(x != father[x]){        father[x] = find(father[x]);    }    return father[x];}int main(){    int n,m;    int i;    while(~scanf("%d%d",&n,&m)){        for(i=0;i<MAXN;i++){            father[i] = i;        }        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        for(i=0;i<m;i++){            scanf("%d%d",&a[i],&b[i]);        }        int sum = 0;        for(i=0;i<m;i++){            int f1 = find(a[i]);            int f2 = find(b[i]);            if(f1 != f2){                if(f1 < f2){                    father[f2] = f1;                }                else{                    father[f1] = f2;                }            }            else{                sum++;            }        }        printf("%d\n",sum);    }    return 0;}