HDOJ-----1856

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

4 1 2 3 4 5 6 1 6 4 1 2 3 4 5 6 7 8

 

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
#include
#include
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
#include
#include
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;
}