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:
    4.net基础之委托事件
    2.net基础之反射
    绕过百度网盘速度限制直接下载百度网盘文件
    1.net基础之泛型
    网页图片按需加载
    小米官网图片轮播
    html+css3实现网页时钟
    接口自动化测试方案详解
    接口测试用例设计实践总结
    Mysql 高可用(MHA)-读写分离(Atlas)
  • 原文地址:https://www.cnblogs.com/pblr/p/4810230.html
Copyright © 2011-2022 走看看