（兩種解法）a Bug's life(P2492)

兩種解法效率上差不多的，都是才學的，，看的。

解一，這個一開始多寫了一個break就悲劇了，，，，

#include<iostream>
#include<stdlib.h>
#include<math.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<string.h>
#include<stack>
#include<math.h>
#include<string>
#include<stdlib.h>
#include<list>
#include<vector>
#include<map>

using namespace std;

#define N 100010

int f[2*N];
int n,m;

int find(int k)
{
    int i,j;
    if (f[k]!=k)
        f[k]=find(f[k]);
    return f[k];
}

void un(int x,int y)
{
    int i,j,k;
    i=find(x);
    j=find(y);
    if (i!=j)
        f[i]=j;
}

int main()
{
    //freopen("fuck.txt","r",stdin);
    int i,j,k;
    int t;
    scanf("%d",&t);
    int num=0;
    while (t--)
    {
        num++;
        scanf("%d%d",&n,&m);
        for(i=1;i<=2*n;i++)
            f[i]=i;
        bool ans=true;
        while (m--)
        {
            scanf("%d%d",&i,&j);
            if (find(i)==find(j))
            {
                ans=false;
                //break;
            }
            if (find(i)!=find(j+n))
            {
                un(i,j+n);
            }
            if (find(j)!=find(i+n))
            {
                un(j,i+n);
            }
        }
        printf("Scenario #%d:\n",num);
        if (ans)
            printf("No suspicious bugs found!\n\n");
        else
            printf("Suspicious bugs found!\n\n");
    }
}
           

解二：

#include<iostream>
#include<stdlib.h>
#include<math.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<string.h>
#include<stack>
#include<math.h>
#include<string>
#include<stdlib.h>
#include<list>
#include<vector>
#include<map>

using namespace std;

#define N 100010

int f[N];
int r[N];
int n,m;

int find(int k)
{
    int i,j;
    if (f[k]!=k)
        f[k]=find(f[k]);
    return f[k];
}

void un(int x,int y)
{
    int i,j,k;
    i=find(x);
    j=find(y);
    if (i!=j)
        f[i]=j;
}

int main()
{
    //freopen("fuck.txt","r",stdin);
    int i,j,k;
    int t;
    scanf("%d",&t);
    int num=0;
    while (t--)
    {
        num++;
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            f[i]=i,r[i]=0;
        bool ans=true;
        while (m--)
        {
            scanf("%d%d",&i,&j);
            if (r[i]==0&&r[j]==0)         // 都沒有交配過時加入對方
            {
                r[i]=j;
                r[j]=i;
            }
            if (r[i])        //一方交配的對方一定是與現在正在交配的是同性。 把同性的加入一個集合
                un(r[i],j);
            if (r[j])
                un(i,r[j]);
            if (find(i)==find(j))       //發現同性
                ans=false;
        }
        printf("Scenario #%d:\n",num);
        if (ans)
            printf("No suspicious bugs found!\n\n");
        else
            printf("Suspicious bugs found!\n\n");
    }
}
           

A Bug's Life

Time Limit: 10000MS	Memory Limit: 65536K
Total Submissions: 20468	Accepted: 6654

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

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!