hdu 5318 The Goddess Of The Moon 矩阵高速幂

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5318

The Goddess Of The Moon

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 438    Accepted Submission(s): 150

Problem Description

Chang’e （嫦娥） is a well-known character in Chinese ancient mythology. She’s the goddess of the Moon. There are many tales about Chang'e, but there's a well-known story regarding the origin of the Mid-Autumn Moon Festival. In a very distant past, ten suns had
risen together to the heavens, thus causing hardship for the people. The archer Yi shot down nine of them and was given the elixir of immortality as a reward, but he did not consume it as he did not want to gain immortality without his beloved wife Chang'e. 

Yi discovered what had transpired and felt sad, so he displayed the fruits and cakes that his wife Chang'e had liked, and gave sacrifices to her. Now, let’s help Yi to the moon so that he can see his beloved wife. Imagine the earth is a point and the moon is
also a point, there are n kinds of short chains in the earth, each chain is described as a number, we can also take it as a string, the quantity of each kind of chain is infinite. The only condition that a string A connect another string B is there is a suffix
of A , equals a prefix of B, and the length of the suffix(prefix) must bigger than one(just make the joint more stable for security concern), Yi can connect some of the chains to make a long chain so that he can reach the moon, but before he connect the chains,
he wonders that how many different long chains he can make if he choose m chains from the original chains.

Input

The first line is an integer T represent the number of test cases.

Each of the test case begins with two integers n, m. 

(n <= 50, m <= 1e9)

The following line contains n integer numbers describe the n kinds of chains.

All the Integers are less or equal than 1e9.

Output

Output the answer mod 1000000007.

Sample Input

2
10 50
12 1213 1212 1313231 12312413 12312 4123 1231 3 131
5 50
121 123 213 132 321

Sample Output

86814837
797922656

Hint

11 111 is different with 111 11

题意：有n个小楼梯，假设两个楼梯的 前缀等于还有一个的后缀就能够首尾相连，前缀后缀长度要大于等于2。

问m个楼梯组成。有多少种组成方法。

做法：要去重，然后judge  每一个楼梯能不能连，构造出构造矩阵。初始矩阵第一行全为1，然后矩阵高速幂。

#include <cstdio>
#include <algorithm>
#include <stdio.h>
#include <string>
#include <set>
#include <math.h>
#include <string.h>
#include <iostream>
using namespace std;

#define Matr 55 //矩阵大小,注意能小就小   矩阵从1開始   所以Matr 要+1   最大能够100
#define ll __int64
struct mat//矩阵结构体。a表示内容，size大小 矩阵从1開始   但size不用加一
{
    ll a[Matr][Matr];
    mat()//构造函数
    {
        memset(a,0,sizeof(a));
    }
};
int Size=  0 ;
ll mod= 1000000007;

mat multi(mat m1,mat m2)//两个相等矩阵的乘法，对于稀疏矩阵。有0处不用运算的优化
{
    mat ans=mat();
    for(int i=1;i<=Size;i++)
        for(int j=1;j<=Size;j++)
            if(m1.a[i][j])//稀疏矩阵优化
                for(int k=1;k<=Size;k++)
                    ans.a[i][k]=(ans.a[i][k]+m1.a[i][j]*m2.a[j][k])%mod; //i行k列第j项
    return ans;
}

mat quickmulti(mat m,ll n)//二分高速幂
{
    mat ans=mat();
    int i;
    for(i=1;i<=Size;i++)ans.a[i][i]=1;
    while(n)
    {
        if(n&1)ans=multi(m,ans);//奇乘偶子乘 挺好记的.
        m=multi(m,m);
        n>>=1;
    }
    return ans;
}

void print(mat m)//输出矩阵信息，debug用
{
    int i,j;
    printf("%d\n",Size);
    for(i=1;i<=Size;i++)
    {
        for(j=1;j<=Size;j++)
			printf("%d ",m.a[i][j]);
        printf("\n");
    }
}
set<string> my;

string str[60];

int judge(string a,string b)
{
	for(int i=2;i<=a.size()&&i<=b.size();i++)
	{
		int flag=1;
		for(int j=0;j<i;j++)
		{
			if(a[a.size()-i+j]!=b[j])
				flag=0;
		}
		if(flag)
			return 1;
	}
	return 0;
}

int  main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,m;
		scanf("%d%d",&n,&m);
		int kk=0;
		my.clear();
		for(int i=0;i<n;i++)
		{
			string ss;
			cin>>ss;
			if(my.find(ss)==my.end())
			{
				my.insert(ss);
				str[++kk]=ss;
			}
		}
		n=kk;
		if(m==0||n==0)
		{
			printf("0\n");
			continue;
		}
		mat gouzao=mat(),chu=mat();//构造矩阵  初始矩阵  

		for(int i=1;i<=kk;i++)
		{
			for(int j=1;j<=kk;j++)
			{
				if(judge(str[i],str[j]))
				gouzao.a[i][j]=1;
			}
		}

		for(int i=1;i<=kk;i++)
		{
			chu.a[1][i]=1;
		}
		Size=kk;
		chu=multi(chu,quickmulti(gouzao,m-1));

		__int64 ans=0;
		for(int i=1;i<=kk;i++)
		{
			ans=(ans+chu.a[1][i])%mod;
		}

		printf("%I64d\n",ans);
	}
	return 0;
}