zoukankan      html  css  js  c++  java
  • HDOJ 5399 Too Simple


    每个函数都必须是一个排列,经过连续的一段确定函数后数字不能少.

    满足上面的条件的话,仅仅要有一个-1函数特别的排列一下就能够满足要求,剩下的能够任意填

    没有-1的话特判


    Too Simple

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 789    Accepted Submission(s): 267


    Problem Description
    Rhason Cheung had a simple problem, and asked Teacher Mai for help. But Teacher Mai thought this problem was too simple, sometimes naive. So she ask you for help.

    Teacher Mai has m functions f1,f2,,fm:{1,2,,n}{1,2,,n}(that means for all x{1,2,,n},f(x){1,2,,n}). But Rhason only knows some of these functions, and others are unknown.

    She wants to know how many different function series f1,f2,,fm there are that for every i(1in),f1(f2(fm(i)))=i. Two function series f1,f2,,fm and g1,g2,,gm are considered different if and only if there exist i(1im),j(1jn),fi(j)gi(j).
     

    Input
    For each test case, the first lines contains two numbers n,m(1n,m100).

    The following are m lines. In i-th line, there is one number 1 or n space-separated numbers.

    If there is only one number 1, the function fi is unknown. Otherwise the j-th number in the i-th line means fi(j).
     

    Output
    For each test case print the answer modulo 109+7.
     

    Sample Input
    3 3 1 2 3 -1 3 2 1
     

    Sample Output
    1
    Hint
    The order in the function series is determined. What she can do is to assign the values to the unknown functions.
     

    Author
    xudyh
     

    Source
     



    /* ***********************************************
    Author        :CKboss
    Created Time  :2015年08月19日 星期三 10时25分44秒
    File Name     :HDOJ5399.cpp
    ************************************************ */
    
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <string>
    #include <cmath>
    #include <cstdlib>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    
    using namespace std;
    
    typedef long long int LL;
    const LL mod=1e9+7LL;
    const int INF=50000000;
    
    int n,m;
    int f[110][110];
    
    int cn;
    int color[110];
    
    bool COLOR(int x,int c)
    {
    	if(color[x]==c) return false;
    	color[x]=c; cn++;
    	return true;
    }
    
    LL QuickPow(LL a,LL n)
    {
    	LL e=1;
    	while(n)
    	{
    		if(n&1) e=(a*e)%mod;
    		a=(a*a)%mod;
    		n/=2;
    	}
    	return e%mod;
    }
    
    bool used[110];
    
    LL jc(LL n)
    {
    	LL ret=1;
    	for(int i=2;i<=n;i++)
    		ret=(ret*i)%mod;
    	return ret%mod;
    }
    
    bool check_P(int x,int L=1,int R=m)
    {
    	if(used[x]==true) return false;
    	int nx=x;
    	for(int i=R;i>=L;i--)
    	{
    		nx=f[i][nx];
    	}
    	if(nx==x) 
    	{
    		if(used[x]==false)
    		{
    			used[x]=true;
    			return true;
    		}
    		else return false;
    	}
    	return false;
    }
    
    bool check_Range(int L,int R)
    {
    	memset(used,false,sizeof(used));
    	bool flag=true;
    	for(int i=1;i<=n&&flag;i++)
    	{
    		int nx=i;
    		for(int j=R;j>=L;j--)
    		{
    			nx=f[j][nx];
    		}
    		if(used[nx]==true) return false;
    		else used[nx]=true;
    	}
    	return true;
    }
    
    int main()
    {
    	//freopen("in.txt","r",stdin);
    	//freopen("out.txt","w",stdout);
    
    	while(scanf("%d%d",&n,&m)!=EOF)
    	{
    		int nig=0;
    		bool check=true;
    		memset(color,0,sizeof(color));
    
    		for(int i=1;i<=m;i++)
    		{
    			scanf("%d",&f[i][1]);
    			if(f[i][1]==-1) nig++;
    			else
    			{
    				cn=0;
    				check=COLOR(f[i][1],i);
    				for(int j=2;j<=n;j++)
    				{
    					scanf("%d",&f[i][j]);
    					check=COLOR(f[i][j],i);
    				}
    				if(cn!=n) check=false;
    			}
    		}
    
    		if(check==false)
    		{
    			puts("0");
    		}
    		else if(check&&nig==0)
    		{
    			/// tePan
    			memset(used,false,sizeof(used));
    			bool flag=true;
    			for(int i=1;i<=n&&flag;i++)
    			{
    				if(check_P(i)==false) flag=false;
    			}
    			if(flag==true) puts("1");
    			else puts("0");
    		}
    		else
    		{
    			//// (n!^(nig-1))
    			bool flag=true;
    			int Left=INF,Right=-INF;
    			f[m+1][1]=-1;
    			for(int i=1;i<=m+1&&flag;i++)
    			{
    				if(f[i][1]==-1)
    				{
    					if(Left!=INF&&Right!=-INF)
    					{
    						/// check;
    						flag=check_Range(Left,Right);
    					}
    					Left=INF; Right=-INF;
    				}
    				else
    				{
    					Left=min(Left,i); Right=max(Right,i);
    				}
    			}
    			if(flag==false) puts("0");
    			else printf("%lld
    ",QuickPow(jc(n),nig-1)%mod);
    		}
    	}
        
        return 0;
    }
    



  • 相关阅读:
    详解Javascript匿名函数的使用 转载自IT EYE Blog
    漂亮、免费和响应式HTML5网站模板 转
    Truth, Equality and JavaScript
    Ember.js 示例
    UX Pin 一款在线界面设计网站
    关于工资的三个秘密
    Semantic Versioning 如何进行版本管理
    CSSS CSS幻灯片
    oracle 导入数据报600错误
    Android 初步Android文件目录介绍
  • 原文地址:https://www.cnblogs.com/yutingliuyl/p/7002188.html
Copyright © 2011-2022 走看看