Friends
題目描述：
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：
Sample Output：
題目大意：
給出了T個測試樣例子，給了n個人，m對朋友，他要求每一個人都擁有同等數量的在線朋友與離線朋友，要求在m對朋友來分配，有多少種方式，意思就是說這裏有m對朋友，但他們沒有標明是在線還是離線，所以由你來標明，分配即可。
思路分析：
這道題首先要看每個人擁有多少個朋友，如果該朋友數量是奇數，說明無法同等分配，如果是偶數才能繼續分配，這裏有個很簡單的思路把，就是朋友的數量是固定的，但分配朋友對的數量的時候，在線朋友對多一個，離線朋友對就少一個，那麼依次爲條件DFS，看是否所有朋友對數量，能否等於m就即可，就多出一種方法。
AC代碼：
用時：15ms
#include<cstdio>
#include<iostream>
#include<string.h>
using namespace std;
int num[10000];//記錄該人有多少朋友
int Leave[10000];//在線人數
int rise[10000];//離線人數
int sum,flag;//flag用於標記，sum用於方法
int m,n;
struct
{
int n1,n2;
}Lemon[10000];//標記朋友對數量
void LemonDFS(int num1)
{
if(num1 == m+1)//如果剛好等於朋友對數量，說明滿足條件
{
sum++;
return;
}
//記錄朋友對的關係
int LF= Lemon[num1].n1;
int RF= Lemon[num1].n2;
if(rise[LF] && rise[RF])//在線人數分配
{
rise[LF]--;
rise[RF]--;
LemonDFS(num1+1);
rise[LF]++;
rise[RF]++;
}
if(Leave[LF] && Leave[RF])//離線人數分配
{
Leave[LF]--;
Leave[RF]--;
LemonDFS(num1+1);
Leave[LF]++;
Leave[RF]++;
}
return;
}
int main()
{
int T;
cin >> T;
while(T--)
{
memset(Lemon,0,sizeof(Lemon));
memset(Leave,0,sizeof(Leave));
memset(rise,0,sizeof(rise));
memset(num,0,sizeof(num));
//刷新4個數組的數據
sum=0;//記錄 方法數量
cin >> n >> m;//輸入人數量與朋友數量
for(int i=1;i<=m;i++)//開始標明第幾號人有多少個朋友
{
cin >> Lemon[i].n1 >> Lemon[i].n2;
num[Lemon[i].n1]++;
num[Lemon[i].n2]++;
}
flag=0;//標記是否有方法
for(int i=1;i<=n;i++)
{
rise[i]=Leave[i]=num[i]/2;//如果朋友對是偶數才能同等分配，如果是奇數
//不滿足條件，退出循環。
if(num[i]%2!=0)
{
flag=1;
break;
}
}
if(flag)//無方法的時候
{
printf("0\n");
}
else//有方法的時候就 DFS分配
{
LemonDFS(1);
printf("%d\n",sum);
}
}
}