HDU 5305(Friends-暴搜)
Friends
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1717 Accepted Submission(s): 854
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.
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.
2
3 3
1 2
2 3
3 1
4 4
1 2
2 3
3 4
4 1
0
2
pid=5392" target="_blank">5392
pid=5391" target="_blank">5391
暴搜
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100+10)
#define MAXM (100+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,m;
int e[MAXM][2];
int degree[MAXN],totdeg[MAXN];
ll ans;
bool check(int x,int y)
{
return ( ( totdeg[x]||( !degree[x] )) && ( totdeg[y]||( !degree[y] )) ); }
void dfs(int p)
{
if (p==m)
{
For(i,n)
if (i!=e[p][0]&&i!=e[p][1]&°ree[i]) return ;
if (degree[e[p][0]]==degree[e[p][1]]&&abs(degree[e[p][0]])==1) {
ans++;
}
return ;
}
// if (p==m+1)
// {
// ans++;
// return;
// }
int x=e[p][0],y=e[p][1];
totdeg[x]--;totdeg[y]--;
degree[x]++;degree[y]++;
if (check(x,y)) dfs(p+1);
degree[x]-=2;degree[y]-=2;
if (check(x,y)) dfs(p+1);
degree[x]++;degree[y]++;
totdeg[x]++;totdeg[y]++;
}
int main()
{
// freopen("F.in","r",stdin); int T;cin>>T;
while(T--) {
ans=0; MEM(degree) MEM(totdeg)
cin>>n>>m;
For(i,m) scanf("%d%d",&e[i][0],&e[i][1]),totdeg[e[i][0]]++,totdeg[e[i][1]]++; bool flag=0;
For(i,n) if (totdeg[i] & 1) {
flag=1;puts("0");break;
}
if (flag) continue; if (m) dfs(1); else ans=1;
printf("%lld\n",ans);
} return 0;
}
HDU 5305(Friends-暴搜)的更多相关文章
- hdu 1979 剪枝暴搜
Fill the blanks Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)T ...
- HDU 4284 Travel (Folyd预处理+dfs暴搜)
题意:给你一些N个点,M条边,走每条边要花费金钱,然后给出其中必须访问的点,在这些点可以打工,但是需要先拿到证书,只可以打一次,也可以选择不打工之直接经过它.一个人从1号点出发,给出初始金钱,问你能不 ...
- HDU 4620 Fruit Ninja Extreme 暴搜
题目大意:题目就是描述的水果忍者. N表示以下共有 N种切水果的方式. M表示有M个水果需要你切. W表示两次连续连击之间最大的间隔时间. 然后下N行描述的是 N种切发 第一个数字C表示这种切法可以切 ...
- hdu 5952 Counting Cliques 求图中指定大小的团的个数 暴搜
题目链接 题意 给定一个\(n个点,m条边\)的无向图,找出其中大小为\(s\)的完全图个数\((n\leq 100,m\leq 1000,s\leq 10)\). 思路 暴搜. 搜索的时候判断要加进 ...
- HDU - 6185 Covering(暴搜+递推+矩阵快速幂)
Covering Bob's school has a big playground, boys and girls always play games here after school. To p ...
- hdu 4400 离散化+二分+BFS(暴搜剪枝还超时的时候可以借鉴一下)
Mines Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Subm ...
- HDU4403(暴搜)
A very hard Aoshu problem Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & ...
- 【BZOJ-3033】太鼓达人 欧拉图 + 暴搜
3033: 太鼓达人 Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 204 Solved: 154[Submit][Status][Discuss] ...
- c++20701除法(刘汝佳1、2册第七章,暴搜解决)
20701除法 难度级别: B: 编程语言:不限:运行时间限制:1000ms: 运行空间限制:51200KB: 代码长度限制:2000000B 试题描述 输入正整数n,按从小到大的顺序输出所有 ...
随机推荐
- Linux学习笔记之Linux命令
1. blkid 查看当前系统中所有已挂载文件系统的类型
- Docker在Ubuntu16.04上安装
转自:http://blog.51cto.com/collen7788/2047800 1.添加Docker源 sudo apt-get update 2.增加CA证书 sudo apt-get in ...
- C语言调用Python
python模块:demo.py def print_arg(str): print str def add(a,b): print 'a=', a print 'b=', b return a + ...
- text-shadow的用法详解
1.兼容性:text-shadow 和 box-shadow 这两个属性在主流现代浏览器上得到了很好的支持( > Chrome 4.0, > Firefox 3.5, > Safar ...
- windows server 2008 如何查看异常重启日志
下面蓝队网络为大家介绍下windows server 2008 如何查看异常重启日志 开始->管理工具->时间查看器 windows日志->系统 筛选当前日志 选择Kernel-Po ...
- 音视频】5.ffmpeg命令分类与使用
GT其实平时也有一些处理音视频的个人或者亲人需求,熟练使用ffmpeg之后也不要借助图示化软件,一个命令基本可以搞定 G: 熟练使用ffmpeg命令!T :不要死记硬背,看一遍,自己找下规律,敲一遍, ...
- cstring to utf8
char* UnicodeToUtf8(CString unicode) { int len; len = WideCharToMultiByte(CP_UTF8, 0, (LPCWSTR)unico ...
- 00JAVA EE
JAVA EE 三层架构 我们的开发架构一般都是基于两种形式,一种是C/S架构,也就是客户端/服务器,另一种是B/S架构,也就是浏览器服务器.在JavaEE开发中,几乎全都是基于B/S架构的开发.那么 ...
- 04Oracle Database 登陆
Oracle Database 登陆 EM Express Login https://localhost:5500/em/login cmd sqlplus SQL/PLUS system/code ...
- Install Zabbix with Docker
1. mysql -uroot -p -h10.10.0.242 zabbix<schema.sqlEnter password: * ERROR 1709 (HY000) at line 86 ...