zoukankan      html  css  js  c++  java
  • hdu 5305 Friends(dfs)

    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 (1n8) and m (0mn(n1)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 xy 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
     1 #include<cstdio>
     2 #include<cstring>
     3 #include<algorithm>
     4 #include<cmath>
     5 using namespace std;
     6 int ans,n,m;
     7 int v[10],v1[10],v2[10];
     8 struct p
     9 {
    10    int u,v;
    11 };p s[60];
    12 void dfs(int i)
    13 {
    14      if (i==m+1)
    15      {
    16          ans++;
    17          return ;
    18      }
    19      if (v1[s[i].u]&&v1[s[i].v])
    20      {
    21           v1[s[i].u]--;
    22           v1[s[i].v]--;
    23           dfs(i+1);
    24           v1[s[i].u]++;
    25           v1[s[i].v]++;
    26      }
    27      if (v2[s[i].u]&&v2[s[i].v])
    28      {
    29           v2[s[i].u]--;
    30           v2[s[i].v]--;
    31           dfs(i+1);
    32           v2[s[i].u]++;
    33           v2[s[i].v]++;
    34      }
    35 }
    36 int main()
    37 {
    38     int i,j,flag,t;
    39     scanf("%d",&t);
    40     while (t--)
    41     {
    42         scanf("%d%d",&n,&m);
    43         flag=0;
    44         memset(v,0,sizeof(v));
    45         memset(v1,0,sizeof(v1));
    46         memset(v2,0,sizeof(v2));
    47         for (i=1;i<=m;i++)
    48         {
    49            scanf("%d%d",&s[i].u,&s[i].v);
    50            v[s[i].u]++;
    51            v[s[i].v]++;
    52         }
    53         for (i=1;i<=n;i++)
    54         {
    55            if (v[i]%2==1)
    56            {
    57               flag=1;
    58               break;
    59            }
    60            v1[i]=v2[i]=v[i]/2;
    61         }
    62         ans=0;
    63         dfs(1);
    64         printf("%d
    ",ans);
    65     }
    66 }
  • 相关阅读:
    公平锁和非公平锁
    读写锁StampedLock的思想
    线程工作窃取算法
    关于SQL注入的问题以及解决方法
    简单工厂模式、工厂模式和抽象工厂模式
    RestFul的无状态规则详解
    Identity Server 4 中文文档(v1.0.0) 目录
    第3章 支持和规范
    第2章 术语
    第1章 背景
  • 原文地址:https://www.cnblogs.com/pblr/p/4810230.html
Copyright © 2011-2022 走看看