zoukankan      html  css  js  c++  java
  • 种类并查集,TOJ(1706)

    题目链接:http://acm.tju.edu.cn/toj/showp1706.html

    很类似Poj的一道帮派的问题,记得找到的可疑的关系,不要将集合刷新就可以了。

    1706.   A Bug's Life
    Time Limit: 5.0 Seconds   Memory Limit: 65536K
    Total Runs: 1190   Accepted Runs: 360    Multiple test files



    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.



    Source: TUD Programming Contest 2005

    #include <stdio.h>
    
    int father[2010];
    int kind[2010];
    
    int Find_Set(int x)
    {
        if(x!=father[x])
        {
            int tmp = father[x];
            father[x] = Find_Set(father[x]);
            kind[x] = (kind[x]+kind[tmp])%2;
        }
        return father[x];
    }
    
    int main()
    {
        int t;
        int cases=1;
        scanf("%d",&t);
        while(t--)
        {
    
            int n,m;
            scanf("%d%d",&n,&m);
    
            for(int i=1;i<=n;i++)
            {
                father[i] = i;
                kind[i] = 0;
            }
            bool flag = true;
            while(m--)
            {
                int x,y;
                scanf("%d%d",&x,&y);
    
                int fx,fy;
                fx = Find_Set(x);
                fy = Find_Set(y);
                if(fx!=fy)
                {
                    father[fy] = fx;
                    kind[fy] = (kind[x]-kind[y]+1)%2;
                }
                else
                {
                    if(kind[x]==kind[y])
                        flag = false;
                }
            }
            if(flag)
                printf("Scenario #%d:
    No suspicious bugs found!
    
    ",cases++);
            else printf("Scenario #%d:
    Suspicious bugs found!
    
    ",cases++);
        }
        return 0;
    }
  • 相关阅读:
    位运算一些简单的应用
    运算符的优先级
    Vue.js 指南-基础
    Nginx将项目配置在子目录
    JavaScript创建读取cookie代码示例【附:跨域cookie解决办法】
    JavaScript动态加载资源【js|css】示例代码
    Uploadifive 使用教程【结合七牛】
    mac 配置快捷键快速锁屏
    七牛大文件上传同步工具配置
    使用七牛云存储作为图床
  • 原文地址:https://www.cnblogs.com/TreeDream/p/5720947.html
Copyright © 2011-2022 走看看