简单的并查集,把有关系的嘉宾放到一个节点下(集合),选择一个嘉宾作为父亲节点,最后只判断有多少个不同的父亲节点即可。
AC代码:
1 #include <stdio.h> 2 #include <algorithm> 3 #include <iostream> 4 #include <string.h> 5 #include <string> 6 #include <math.h> 7 #include <stdlib.h> 8 #include <queue> 9 #include <stack> 10 #include <set> 11 #include <map> 12 #include <list> 13 #include <iomanip> 14 #include <vector> 15 #pragma comment(linker, "/STACK:1024000000,1024000000") 16 #pragma warning(disable:4786) 17 18 using namespace std; 19 20 const int INF = 0x3f3f3f3f; 21 const int MAX = 1000 + 10; 22 const double eps = 1e-8; 23 const double PI = acos(-1.0); 24 25 int a[MAX]; 26 27 int find(int x) 28 { 29 while(x != a[x]) 30 x = a[x]; 31 return x; 32 } 33 34 int main() 35 { 36 int T; 37 scanf("%d",&T); 38 { 39 while(T --) 40 { 41 int n , m; 42 scanf("%d %d", &n , &m); 43 int i; 44 for(i = 0;i <= n;i ++) 45 a[i] = i; 46 int temp1 , temp2; 47 while(m --) 48 { 49 scanf("%d %d",&temp1 , &temp2); 50 int fa = find(temp1); 51 int fb = find(temp2); 52 if(fa != fb) 53 a[fa] = fb; 54 } 55 int ans = 0; 56 for(i = 1;i <= n;i ++) 57 if(a[i] == i) 58 ans ++; 59 printf("%d\n",ans); 60 } 61 } 62 return 0; 63 }
原文:http://www.cnblogs.com/jeff-wgc/p/4449360.html