hdoj1213和zoj2833 （简单并查集）

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hdoj1213

最典型的并查集应用，如果节点a的祖先节点和节点b的祖先节点不相同，则将它们合并

#include <stdio.h>
#include <stdlib.h> 

int root[10000];
int find(int i){
    return root[i] == i ? i : find(root[i]);
}

int main(){
    int m, n, t;
    int a, b, i;
    int cnt, x, y;
    scanf("%d", &t);
    while(t--){
        cnt = 0;
        scanf("%d%d", &m, &n);
        
        for(i = 1; i <= m; ++i) root[i] = i;
        for(i = 0; i < n; ++i){
            scanf("%d%d", &a, &b);
            x = find(root[a]);
            y = find(root[b]);
            if(x != y) root[x] = y;
        }
        
        for(i = 1; i <= m; ++i){
            if(root[i] == i) cnt++;
        }
        printf("%d\n", cnt);
    }
    return 0;
}

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zoj2833
这是一个并查集的应用，当两个分量合并时刷新伙伴个数，另外需要注意一下换行符和优化问题。

#include<stdio.h>
#include<stdlib.h>

const int maxn = 100000+500;
int root[maxn];
int ans[maxn];

int find(int i){
      return root[i] == i ? i : find(root[i]); 
}

int main(){
       int a, b, m, n, x, y;
       int i, j = 0;
       int t, k;
       char ch;
       while(scanf("%d%d", &m, &n) != EOF){ 
             j++;
             if(j != 1) printf("\n");
             printf("Case %d:\n", j);
             for(i = 1; i <= m; i++)  {root[i] = i; ans[i] = 1;}
             for(i = 0; i <n; i++){     
                   getchar();
                   scanf("%c", &ch);
                   if(ch == 'M'){      
                         scanf("%d%d", &a, &b);
                         x = find(a);
                         y = find(b);
                         root[a] = x;
                         root[b] = y;
                         if(x != y)  { root[x] = y; ans[y] += ans[x];}
                   }
                   if(ch == 'Q') {      
                          scanf("%d", &t);
                          printf("%d\n", ans[find(t)]);
                   }
             }
       }
       return 0;
}