hdu 5318 2015多校对抗赛三

The Goddess Of The Moon

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 842    Accepted Submission(s): 373

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

210 5012 1213 1212 1313231 12312413 12312 4123 1231 3 1315 50121 123 213 132 321

 

Sample Output

86814837797922656
Hint
11 111 is different with 111 11
 

 

题意：有n段小梯子吧，要选m段组成一个大的长的梯子，问有多少种不同的选法。m的范围很大。

为了固定梯子，要求两段梯子要有至少2长的公共子串，A的后缀是B的前缀。那么长度为1的字段梯子都可以丢掉了。

坑点：输入有重复的梯子，需要去重。

做法：dp[m][i]表示长度为m，当前最后段为第i段字段的种数。那么转移方程为：

dp[m][i]=sigma( dp[m-1][j] ) 当i-j能连边

m的规模很大， 转移会T。

类似斐波那契数列第n项，可以用矩阵快速幂转移。

矩阵为各字段能否连接的关系

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long long#define MOD 1000000007struct duan{    char name[100];}p[60];struct mtx{    LL f[60][60];    LL n;    mtx(LL n_=0):n(n_){memset(f,0,sizeof f); }    mtx operator * (mtx b){        mtx c(n),a=*this;        for(int i=0;i<n;i++)        for(int j=0;j<n;j++){            c.f[i][j]=0;            for(int k=0;k<n;k++)                c.f[i][j]=(c.f[i][j]+a.f[i][k]*b.f[k][j])%MOD;        }        return c;    }    mtx operator ^ (LL k){        mtx c(n),a=*this;        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)                c.f[i][j]=i==j?1:0;        while(k){            if(k&1)c=c*a;            a=a*a;            k>>=1;        }        return c;    }};char tmp1[100],tmp2[100];bool judge(char *s1,char *s2){    int len1=strlen(s1);    int len2=strlen(s2);    int len=min(len1,len2);    int t=0;    tmp1[t]=s1[--len1];    tmp2[t]=s2[t++];    while(t+1<=len){        tmp1[t]=s1[--len1];        tmp2[t]=s2[t++];        bool eq=true;        for(int i=0;i<t;i++){            if(tmp1[t-1-i]!=tmp2[i]){                eq=false;                break;            }        }        if(eq)return true;    }    return false;}int main(){//    freopen("5318.in","r",stdin);//    freopen("5318.out","w",stdout);    int T;scanf("%d",&T);    while(T--){        LL n,m;scanf("%lld%lld",&n,&m);        LL cnt=0;        while(n--){            scanf("%s",p[cnt].name);            if(strlen(p[cnt].name)<=1)continue;            bool have=false;            for(int j=0;j<cnt;j++){                if(strcmp(p[cnt].name,p[j].name)==0){                    have=true;                    break;                }            }            if(have==false)cnt++;        }        n=cnt;        mtx ans(n);        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)                if(judge(p[i].name,p[j].name))                    ans.f[i][j]=1;        if(m<1){puts("0");continue; }        ans=ans^(m-1);        LL tot=0;        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)                tot=(tot+ans.f[i][j])%MOD;        printf("%lld\n",tot);    }    return 0;}