A Bug's Life(向量偏移)

A Bug's Life

Time Limit : 15000/5000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 97   Accepted Submission(s) : 38

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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

2
3 3
1 2
2 3
1 3
4 2
1 2
3 4

Sample OutpuScenario #1、

Suspicious bugs foundScenario #2:

No suspicious bugs found!

http://blog.csdn.net/shuangde800/article/details/7974664
这位大佬写的向量偏移非常阅读后这题就非常容易了；
就是他写的方程有点奇怪；
我自己觉得是这样的：
aa->bb=(-delta[a]+d-1+delta[b])%3;
a->b=(delta[a]-delta[b])%3;


这题题意用一组数据来说明吧
1 2；2 3；
1 3

1和2是异性，2和3也是异性，也就是1和3是同性，但是他有将a 和b是异性。出现了bug；

#include <cstring>
#include<stdio.h>
int p[55000];
int f[55000];//i的根节点到i的偏移量
int flag;

int findi(int x)
{
    if(p[x]==x)
        return x;
    int t=p[x];
    p[x]=findi(p[x]);
    f[x]=(f[x]+f[t])%2;
    return p[x];
}
void unioni(int x,int y)
{
    int fx=findi(x);
    int fy=findi(y);
    p[fx]=fy;
    f[fx]=(f[y]-f[x]+1)%2;
    return ;
}

int main (){
    int t;

    scanf("%d",&t);
    int k=1;
    while(t--)
        {
        flag=0;
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;++i){
            p[i]=i;
            f[i]=0;
        }
        int a,b;
        while(m--){
            scanf("%d%d",&a,&b);
            int fa=findi(a);
            int fb=findi(b);
            if(fa==fb){
                if(f[a]==f[b])
                flag=1;
            }
            else unioni(a,b);
        }
        if(flag)
             printf("Scenario #%d:
Suspicious bugs found!

",k++);
        else
             printf("Scenario #%d:
No suspicious bugs found!

",k++);
    }
    return 0;
}