HDOJ-----2120并查集（水题）

最新推荐文章于 2019-10-20 14:31:55 发布

pupil_blue

最新推荐文章于 2019-10-20 14:31:55 发布

阅读量187

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/blue_hpu/article/details/52088788

本文介绍了一个利用并查集算法解决的问题：给定一系列边，判断能够形成多少个互不相交的环状区域。通过并查集进行节点合并与检测，最终输出最多可以独立分配给勤奋参赛者的土地数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1270 Accepted Submission(s): 761

Problem Description

ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

Input

In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.

Output

Output the maximum number of ACMers who will be awarded.
One answer one line.

Sample Input

Sample Output

看英文题就是头疼，看懂了原来就是个这

就是给一堆边，问能围成多少环，并查集检验一下就行了

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 1100 
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;
        return ;
    }
    ans++;
}
int main(){
    int x, y, t, m, n;
    while(~scanf("%d%d", &x, &y)){
        for(int i = 0; i < x; i++){
            pre[i] = i;
        }
        ans = 0;
        for(int i = 0; i < y; i++){
            scanf("%d%d", &m, &n);
            merge(m, n);
        }
        printf("%d\n", ans);
    }
    return 0;
}