hdu 1856 More is better

来源：互联网发布：阿里云自定义端口编辑：程序博客网时间：2024/05/22 13:30

点击hdu 1856思路:
思路: 离散化+并查集
分析:
1 点数最多为10^7，但是边数最多10^5，所以我们必须采用离散化，然后利用带权并查集的思想，rank[x]表示的是以x为根节点的集合的元素个数
2 这一题主要注意的就是当n = 0的时候，因为题目说了刚开始有10^7个人在房间里面，所以n = 0的时候最多有一个人出去

代码:

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAXN = 200010;struct Node{    int x;    int y;};Node node[MAXN];int n , pos;int num[MAXN];int father[MAXN];int rank[MAXN];void init(){    sort(num , num+pos);    pos = unique(num , num+pos)-num;    for(int i = 0 ; i <= pos ; i++){       father[i] = i;       rank[i] = 1;    }}int search(int x){    int left = 0;    int right = pos-1;    while(left <= right){         int mid = (left+right)>>1;         if(num[mid] == x)             return mid+1;         else if(num[mid] < x)             left = mid+1;         else             right = mid-1;    }}int find(int x){    if(father[x] != x)        father[x] = find(father[x]);    return father[x];}int solve(){    init();    int ans = 0;    for(int i = 0 ; i < n ; i++){        int x = search(node[i].x);         int y = search(node[i].y);         int fx = find(x);        int fy = find(y);        if(fx != fy){            father[fx] = fy;             rank[fy] += rank[fx];            ans = max(ans , rank[fy]);        }    }    return n == 0 ? 1 : ans;}int main(){    while(scanf("%d" , &n) != EOF){         pos = 0;         for(int i = 0 ; i < n ; i++){            scanf("%d%d" , &node[i].x , &node[i].y);                 num[pos++] = node[i].x;            num[pos++] = node[i].y;         }         printf("%d\n" , solve());    }    return 0;}