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The Goddess Of The Moon Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 746    Accepted Submission(s): 326 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.  However, while Yi went out hunting, Fengmeng broke into his house and forced Chang‘e to give up the elixir of immortality to him, but she refused to do so. Instead, Chang‘e drank it and flew upwards towards the heavens, choosing the moon as residence to be nearby her beloved husband. 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   Source 2015 Multi-University Training Contest 3

#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<map>
#include<set>
#include<time.h>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1e-8
#define INF 0x3f3f3f3f
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define MOD 1000000007
#define maxn 55
int n, m;

struct Matrix
{
    LL go[maxn][maxn];
    Matrix()
    {
        memset(go, 0, sizeof(go));
    }
};

Matrix mul(Matrix a, Matrix b)
{
    Matrix tmp;
    for(int k=1; k<=n; k++)
            for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                        tmp.go[i][j] = (tmp.go[i][j] + a.go[i][k]*b.go[k][j]) % MOD;
    return tmp;
}

Matrix mul_pow(Matrix a, int k)
{
    Matrix res;
    for(int i=1; i<=n; i++)
    res.go[1][i] = 1;
    while(k)
    {
        if(k & 1)
        res = mul(res, a);
        a = mul(a, a);
        k >>= 1;
    }
    return res;
}




set<string> S;
string str[maxn];
int check(string a, string b)
{
    int len1 = a.size(), len2 = b.size();
    if(len1 == 1 || len2 == 1)
        return 0;
    for(int i=len1-2; i>=0; i--)
    {
        int j = 0;
        int t = i;
        while(a[t] == b[j] && j < len2 &&  t < len1)
        {
            j++;
            t++;
        }
        if(t == len1)
            return 1;
    }
    return 0;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        S.clear();
        scanf("%d%d", &n, &m);
        for(int i=0; i<n; i++)
        {
            cin>>str[i];
            S.insert(str[i]);
        }
        n = S.size();
        int cnt = 0;
        for(set<string>::iterator it=S.begin(); it!=S.end(); it++)
            str[cnt++] = *it;
        Matrix M;
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
            {
                if(check(str[i], str[j]))
                    M.go[i+1][j+1] = 1;
            }

        Matrix ans, tmp = M;
        tmp = mul_pow(tmp, m-1);
        for(int i=1; i<=n; i++)
            ans.go[1][i] = 1;

        ans = mul(ans, tmp);
//        m--;
//        while(m)
//        {
//            if(m & 1)
//                ans = mul(ans , tmp);
//            tmp = mul(tmp , tmp);
//            m >>= 1;
//        }
        LL sum = 0;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                sum = (sum + ans.go[i][j]) % MOD;
        printf("%I64d\n", sum);
    }
    return 0;
}

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标签：acm   hdu   

原文地址：http://blog.csdn.net/dojintian/article/details/47145227