zoukankan      html  css  js  c++  java
  • hdu 1856 More is better(并查集)

    题意:

    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.

    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)

    思路:

    简单并查集

    代码:

    int n;
    struct node{
        int x,y;
    }
    Pair[100005];
    
    int cn;
    int a[200005];
    int ct[200005];
    int fa[200005];
    
    int findFa(int x){
        return fa[x]==x?fa[x]:fa[x]=findFa(fa[x]);
    }
    
    void Union(int x,int y){
        int fx=findFa(x);
        int fy=findFa(y);
        if(fx!=fy){
            fa[fx]=fy;
        }
    }
    
    
    
    int main(){
        map<int,int> mp;
    
        while(scanf("%d",&n)!=EOF){
            mp.clear();
            cn=0;
            rep(i,1,n){
                scanf("%d%d",&Pair[i].x,&Pair[i].y);
                if(mp[Pair[i].x]==0){
                    mp[Pair[i].x]=1;
                    a[++cn]=Pair[i].x;
                }
                if(mp[Pair[i].y]==0){
                    mp[Pair[i].y]=1;
                    a[++cn]=Pair[i].y;
                }
            }
            sort(a+1,a+1+cn);
            rep(i,1,cn) fa[i]=i;
            rep(i,1,n){
                int px=lower_bound(a+1,a+1+cn,Pair[i].x)-a;
                int py=lower_bound(a+1,a+1+cn,Pair[i].y)-a;
                Union(px,py);
            }
            mem(ct,0);
            int ans=1;
            rep(i,1,cn){
                int t=findFa(i);
                ct[t]++;
                ans=max( ans,ct[t] );
            }
            printf("%d
    ",ans);
        }
    
        return 0;
    }
  • 相关阅读:
    配置JDK
    360首页(练习)
    表单练习——(简单的注册页面)
    主页
    证明某字母是否最后一个字母
    方法的重载与重写区别
    什么是设计模式
    java 静态方法和实例方法的区别
    什么是静态方法
    手机充电(练习)
  • 原文地址:https://www.cnblogs.com/fish7/p/4334477.html
Copyright © 2011-2022 走看看