A Bug's Life POJ - 2492 （种类并查集）

最新推荐文章于 2022-04-03 23:46:16 发布

原创最新推荐文章于 2022-04-03 23:46:16 发布 · 654 阅读

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本文解析了一道关于稀疏图算法的竞赛题目，通过使用并查集数据结构，解决了一个关于昆虫交互行为的问题，判断是否存在违反两性假设的交互。文章详细介绍了算法思路，包括初始化并查集、定义根节点查找和合并操作，以及如何根据输入数据进行判断。

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Background
Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs.
Problem
Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.
Input
The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.
Output
The output for every scenario is a line containing “Scenario #i:”, where i is the number of the scenario starting at 1, followed by one line saying either “No suspicious bugs found!” if the experiment is consistent with his assumption about the bugs’ sexual behavior, or “Suspicious bugs found!” if Professor Hopper’s assumption is definitely wrong.
Sample Input
2
3 3
1 2
2 3
1 3
4 2
1 2
3 4
Sample Output
Scenario #1:
Suspicious bugs found!

Scenario #2:
No suspicious bugs found!
Hint
Huge input,scanf is recommended.
题解：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAXN 1000000
int pre[MAXN*2];//开一个大一点的数组，前面的存储一个性别的，后面的再存储一个性别的
int root(int a)//寻找根节点
{
    while(a!=pre[a])
        a = pre[a];
    return a;
}
void merge(int a, int b)//合并
{
    int x = root(a);
    int y = root(b);
    if(x!=y)
        pre[x] = y;
}
int main()
{
    int t, n, m;
    scanf("%d", &t);//组数
    for(int i=1;i<=t;i++)
    {
        for(int k=1;k<=2*MAXN;k++)
            pre[k] = k;//初始化
        int flag = 1;//标记
        scanf("%d %d", &n, &m);
        for(int j=1;j<=m;j++)
        {
          int x, y;
          scanf("%d %d", &x, &y);
        if(flag==0)//已经找到性别相同的
            continue;
        if(root(x)==root(y)||root(x+MAXN)==root(y+MAXN))//根节点一样，性别相同
        {
            flag = 0;//标记
            continue;
        }
        else
        {
            //合并
            merge(x, y+n);
            merge(x+n, y);
        }
        }
        printf("Scenario #%d:\n", i);
        if(flag==0)
            printf("Suspicious bugs found!\n\n");//输出空行
        else
            printf("No suspicious bugs found!\n\n");
    }
    return 0;
}