hdu5305Friends

题意：给出n个人，m对朋友，要求每个人的A类朋友跟B类朋友一样多，求种类数

……不会做……我想爆搜……当时我还太年轻，并不知道怎么爆搜……

这样，维护一个d数组，di表示i的状态，若i有一个A就+1，否则-1，这样若是一样多，肯定di最后=0

爆搜所有边的状态即可，因为边足够多的时候，有些边之间相互制约，所以实际上跑得并不太慢

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
bool edge[10][10];
int d[10],ans,n,m;
void dfs(int v,int u)
{
if(v>n)
{
ans++;
return;
}
if(u>n)
{
if(d[v])
return;
dfs(v+1,v+2);
return;
}
if(edge[v][u])
{
d[v]++;
d[u]++;
dfs(v,u+1);
d[v]-=2;
d[u]-=2;
dfs(v,u+1);
d[v]++;
d[u]++;
}
else
dfs(v,u+1);
}
int main()
{
int T;
cin>>T;
while(T--)
{
cin>>n>>m;
memset(edge,0,sizeof(edge));
while(m--)
{
int a,b;
cin>>a>>b;
edge[a]=edge[b][a]=1;
}
ans=0;
memset(d,0,sizeof(d));
dfs(1,2);
cout<<ans<<endl;
}
}

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 731    Accepted Submission(s): 346


Problem Description

There are n people
and m pairs
of friends. For every pair of friends, they can choose to become online friends (communicating using online applications) or offline friends (mostly using face-to-face communication). However, everyone in these n people
wants to have the same number of online and offline friends (i.e. If one person has x onine
friends, he or she must have x offline
friends too, but different people can have different number of online or offline friends). Please determine how many ways there are to satisfy their requirements. 

 

Input

The first line of the input is a single integer T (T=100),
indicating the number of testcases. 

For each testcase, the first line contains two integers n (1≤n≤8) and m (0≤m≤n(n−1)2),
indicating the number of people and the number of pairs of friends, respectively. Each of the next m lines
contains two numbers x and y,
which mean x and y are
friends. It is guaranteed that x≠y and
every friend relationship will appear at most once. 

 

Output

For each testcase, print one number indicating the answer.

 

Sample Input

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

 

Sample Output

0
2

 

Source

2015 Multi-University Training Contest 2