hdu 5318 The Goddess Of The Moon 【矩阵快速幂】

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

题意：
给你n种串。，每种无限个，选择m个物品，问你可以组成几种串。
如果a串的后缀和b串的前缀相等，并且长度>=2，则b串可以连在a串后面（注意，不用合并a，b串相同的位置）。

思路：输入的字符串可能有相同的，注意去重。
若s[i]，s[j]可以链接，则ok[i][j] = 1 ; 否则为 0 ；

定义dp[i][j] 表示选择i个串，一、以j串结尾的方案数。
则dp[i][j] = dp[i-1][k]*ok[k][j] (1<= k <= n)

res矩阵第一行初使为1，其他为 0 .

代码：

#include <stdio.h>#include <ctime>#include <math.h>#include <limits.h>#include <complex>#include <string>#include <functional>#include <iterator>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <list>#include <bitset>#include <sstream>#include <iomanip>#include <fstream>#include <iostream>#include <ctime>#include <cmath>#include <cstring>#include <cstdio>#include <time.h>#include <ctype.h>#include <string.h>#include <string>#include <assert.h>using namespace std;int n, m;const int MOD = 1000000007;struct node{    int ok[55][55];}is;node multi(node a, node b){    node tmp;    memset(tmp.ok,0,sizeof(tmp.ok));    for (int i = 1; i <= n; i++)        for (int j = 1; j <= n; j++)            for (int k = 1; k <= n; k++)            {                tmp.ok[i][j] += ((long long)a.ok[i][k] * (long long)b.ok[k][j]) % MOD;                tmp.ok[i][j] %= MOD;            }    return tmp;}//////////////////char s[100][20];bool check(int x, int y) //判断是否符合前缀 后缀要求{    int lenx = strlen(s[x]), leny = strlen(s[y]);    if (lenx == 1 || leny == 1) return false;    for (int i = lenx - 2; i >= 0; i--)     {        int j = 0, ti = i;        while (ti < lenx && j < leny && s[x][ti] == s[y][j]) ti++, j++;        if (ti == lenx) return true;    }    return false;}void solve(){    int tm = m;    node tmp = is, res;    memset(res.ok, 0, sizeof(res.ok));    for (int i = 1; i <= n; i++) res.ok[1][i] = 1;    tm--;    while (tm)    {        if (tm & 1)            res = multi(res,tmp);        tmp = multi(tmp, tmp);        tm >>= 1;    }    int ans = 0;    for (int i = 1; i <= n; i++)        ans = (ans + res.ok[1][i]) % MOD;    printf("%d\n",ans);}set<string> vec;int main(){    int t;    scanf("%d",&t);    while (t--)    {        memset(is.ok, 0, sizeof(is.ok));        vec.clear();        scanf("%d%d", &n, &m);        for (int i = 1; i <= n; i++)        {            scanf("%s",s[i]);            vec.insert(s[i]);        }        int num = 0;        for (auto &it : vec) strcpy(s[++num],it.c_str());        n = num;        for (int i = 1; i <= n; i++)            for (int j = 1; j <= n; j++)                if (check(i, j)) is.ok[i][j] = 1;        solve();    }    return 0;}