【冰茶几专题】F - A Bug's Life

最新推荐文章于 2023-02-27 19:21:17 发布

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最新推荐文章于 2023-02-27 19:21:17 发布

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原文链接：http://www.cnblogs.com/111qqz/p/4436549.html

本文介绍了一个算法问题，通过处理虫子间的交互数据来验证一种关于虫子性别的假设。输入包含多个场景，每个场景列出一定数量的虫子及其交互情况。算法需判断这些交互是否符合只有两种性别且仅异性间相互作用的假设。

F - A Bug's Life

Time Limit:10000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Description

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

Sample Output

Scenario #1:
Suspicious bugs found!

Scenario #2:
No suspicious bugs found!

Hint

Huge input,scanf is recommended.

也是带种类的冰茶几。

由于只分了两类...我们还是可以按照上道题的做法。。

感觉完全是一样的题啊。。

结果一直WA。。。。

最后发现。。。我边读入边判断。。发现同性恋了就直接Break掉了。。。后面改组的数据读到下一组去了233，不WA就日了汪了。。。

还是把数据的读完再进行操作比较好==

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;

const int N=2E5+7;
bool flag;
int f[N];
int T,n,m,x,y;
char ch;

int root (int x)
{
    if (f[x]!=x)
        f[x] = root(f[x]);
    return f[x];
}
void u(int a,int b)
{
    f[root(a)]=root(b);
}
int main()
{
    scanf("%d",&T);
    for ( int cas = 1 ; cas <= T ; cas++ )
    {
        for ( int i = 1 ; i < N ; i++ )
            f[i] = i;
        scanf("%d %d",&n,&m);
        flag = false;
        for ( int i = 1 ; i <= m ; i++ )
        {

            scanf("%d %d",&x,&y);
            if (flag)
                continue;

            if (root(x)==root(y)||root(x+n)==root(y+n))
            {

                flag = true;
            }
            u(x,y+n);
            u(x+n,y);
        }
        if ( !flag )
        {
            printf("Scenario #%d:\n",cas);
            printf("No suspicious bugs found!\n");

        }
        else
        {
                printf("Scenario #%d:\n",cas);
                printf("Suspicious bugs found!\n");

        }
        cout<<endl;


    }






    return 0;
}