hdu 1829 A Bug's Life （判断二分图）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1829

A Bug's Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 12112    Accepted Submission(s): 3952

Problem 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

23 31 22 31 34 21 23 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, Darmstadt, Germany

 

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题目大意：给出一些关系，判断是否存在同性恋，简单的说就是判断是否是二分图。

这里新学了一种建图的方法，离散化建图。

解题思路：二分图的判断+bfs，如果不用离散化建图会数组越界RE。

详见代码。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;int judge[2010];int n,m;int num,head[2010];struct edge{    int st,ed,next;} e[2010*1000];void addedge(int x,int y){    e[num].st=x;    e[num].ed=y;    e[num].next=head[x];    head[x]=num++;    e[num].st=y;    e[num].ed=x;    e[num].next=head[y];    head[y]=num++;}bool bfs(int x){    queue<int>q;    q.push(x);    judge[x]=0;    //cout<<x<<endl;    while (!q.empty())    {        int s=q.front();        q.pop();        for(int i=head[s]; i!=-1; i=e[i].next)        {            //cout<<judge[i]            if (judge[e[i].ed]==-1)            {                judge[e[i].ed]=(judge[s]+1)%2;                q.push(e[i].ed);            }            else            {                if (judge[e[i].ed]==judge[s])                    return false;            }        }    }    return true;}int main(){    int flag;    int tt=1;    int t;    scanf("%d",&t);    while (t--)    {        scanf("%d%d",&n,&m);        flag=1;        //memset(Map,0,sizeof(Map));        memset(judge,-1,sizeof(judge));        int a,b;        num=0;        memset(head,-1,sizeof(head));        for (int i=1; i<=m; i++)        {            scanf("%d%d",&a,&b);            addedge(a,b);        }        for (int i=1; i<=n; i++)        {            if(judge[i]==-1)            {                if (!bfs(i))                    flag=0;            }        }        printf ("Scenario #%d:\n",tt++);        if (flag==1)            printf ("No suspicious bugs found!\n");        else            printf ("Suspicious bugs found!\n");        printf ("\n");    }    return 0;}