zoukankan      html  css  js  c++  java
  • 2015暑假多校联合---Friends(dfs枚举)

    原题链接

    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
     
    Author
    XJZX
     
    Source
     
    Recommend
    wange2014
     
    题意:输入n,m,n表示有n个人,m表示m对朋友关系,现在要使每个人的朋友划分为在线朋友和离线朋友,且在线朋友和离线朋友数量相等(一对朋友之间只能是在线朋友或者离线朋友),求方案数;
     
    思路:用dfs深搜枚举每一条边(即每一对朋友关系),若能深搜进行完最后一条边,即当前边cnt==m+1  则ans++;
     
    代码如下:
    #include <iostream>
    #include <algorithm>
    #include <queue>
    #include <vector>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    int n,m,cnt,ans;
    int c1[10],c2[10],d[10];
    struct Node
    {
        int u,v;
    }node[200];
    
    void dfs(int i)
    {
        if(i-1==m)
        {
            ans++;
            return ;
        }
        if(c1[node[i].u]&&c1[node[i].v])
        {
            c1[node[i].u]--;
            c1[node[i].v]--;
            dfs(i+1);
            c1[node[i].u]++;
            c1[node[i].v]++;
        }
        if(c2[node[i].u]&&c2[node[i].v])
        {
            c2[node[i].u]--;
            c2[node[i].v]--;
            dfs(i+1);
            c2[node[i].u]++;
            c2[node[i].v]++;
        }
    }
    
    int main()
    {
        int T;
        cin>>T;
        while(T--)
        {
            cnt=0;
            ans=0;
            scanf("%d%d",&n,&m);
            memset(node,0,sizeof(node));
            memset(c1,0,sizeof(c1));
            memset(c2,0,sizeof(c2));
            memset(d,0,sizeof(d));
            for(int i=1;i<=m;i++)
            {
                int u,v;
                scanf("%d%d",&u,&v);
                node[++cnt].u=u;
                node[cnt].v=v;
                d[u]++;
                d[v]++;
            }
            int flag=0;
            for(int i=1;i<=n;i++)
            {
                c1[i]=c2[i]=d[i]/2;
                if(d[i]&1)
                {
                    flag=1;
                    break;
                }
            }
            if(flag)
            {
                puts("0");
                continue;
            }
            dfs(1);
            printf("%d
    ",ans);
        }
        return 0;
    }
     
  • 相关阅读:
    心得体会,搞清楚你为什么学习C++?
    完整版本的推箱子小游戏,最简单的纯C语言打造
    联合体、枚举体初步了解及运用
    结构体的初步了解
    使用 Appium 测试微信小程序 Webview——打开调试功能
    Jmeter 使用ssh command 链接linux
    jmeter响应内容乱码问题
    Mac 更新 node版本
    解决jenkins + ant + jmeter发送邮件失败的问题:java.lang.ClassNotFoundException: javax.mail.internet.MimeMessage
    bash特殊字符-2
  • 原文地址:https://www.cnblogs.com/chen9510/p/5813781.html
Copyright © 2011-2022 走看看