HDOJ-----1856

最新推荐文章于 2019-07-25 09:33:17 发布

原创最新推荐文章于 2019-07-25 09:33:17 发布 · 245 阅读

0 ·

CC 4.0 BY-SA版权

并查集专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一个基于并查集算法的问题解决方法，旨在找出在给定直接朋友关系对的情况下，能够形成的最大朋友圈规模。通过实现并优化算法，确保了内存使用效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 24376 Accepted Submission(s): 8752

Problem Description

Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements.

Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

Input

The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)

Output

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

Sample Input

Sample Output

4
2

带权并查集，就是问最多的一棵树有多少元素，和传染病那个差不多

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 10000010
int pre[maxn], dis[maxn];
int ok, ans;
int find(int a){
    return a == pre[a] ? a : pre[a] = find(pre[a]);
}
void merge(int x, int y){
    int fx = find(x), fy = find(y);
    if(fx != fy){
        pre[fx] = fy;
        dis[fy] += dis[fx];
    }
}
int main(){
    int x, y, t;
    while(scanf("%d", &t) != EOF){
        ans = 0;
        for(int i = 1; i < maxn; i++){
            pre[i] = i;
            dis[i] = 1;
        }
        for(int i = 0; i < t; i++){
            scanf("%d%d", &x, &y);
            merge(x, y);
            ans = max(ans, max(x, y));
        }
        if(t == 0){
            printf("1\n");
            continue;
        }
        int cnt = 0;
        for(int i = 1; i <= ans; i++){
            if(cnt < dis[i]){
                cnt = dis[i];
            }
        }
        printf("%d\n", cnt);
    }
    return 0;
}

上边的占内存太大，优化了一下

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 100005 
int pre[maxn], dis[maxn];
int ok, ans;
int find(int a){
    return a == pre[a] ? a : pre[a] = find(pre[a]);
}
void merge(int x, int y){
    int fx = find(x), fy = find(y);
    if(fx != fy){
        pre[fx] = fy;
        dis[fy] += dis[fx];
    }
    ans = ans > dis[fy] ? ans : dis[fy];
}
int main(){
    int x, y, t;
    while(~scanf("%d", &t)){
        ans = 0;
        memset(dis, 0, sizeof(dis));
        memset(pre, 0, sizeof(pre));
        for(int i = 0; i < t; i++){
            scanf("%d%d", &x, &y);
            if(!dis[x]){
                pre[x] = x;
                dis[x] = 1;
            }
            if(!dis[y]){
                pre[y] = y;
                dis[y] = 1;
            }
            merge(x, y);
        }
        if(t == 0){
            printf("1\n");
            continue;
        }
        printf("%d\n", ans);
    }
    return 0;
}