ZOJ 3644 Kitty's Game（DP)

Description

Kitty is a little cat. She is crazy about a game recently.

There are n scenes in the game(mark from 1 to n). Each scene has a number p_i. Kitty's score will become least_common_multiple(x,p_i) when Kitty enter the i^th scene. x is the score that Kitty had previous. Notice that Kitty will become mad If she go to another scene but the score didn't change.

Kitty is staying in the first scene now(with p₁ score). Please find out how many paths which can arrive at the n^th scene and has k scores at there. Of course, you can't make Kitty mad.

We regard two paths different if and only if the edge sequence is different.

Input

There are multiple test cases. For each test case:

The first line contains three integer n(2 ≤ n ≤ 2000), m(2 ≤ m ≤ 20000), k(2 ≤ k ≤ 10⁶). Then followed by m lines. Each line contains two integer u, v(1 ≤ u, v ≤ n, u ≠ v) indicate we can go to v^th scene from u^th scene directly. The last line of each case contains n integer p_i(1 ≤p_i ≤ 10⁶).

Process to the end of input.

Output

One line for each case. The number of paths module 1000000007.

Sample Input

5 6 841 22 51 33 51 44 51 5 4 12 21

Sample Output

2

题意：n：n个点 m：m条路  k：值，每个点有一个值，要求从起点到终点的不同方案。中间不用许出现LCM不变的情况

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<map>typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int maxn=2010;const int MOD=1e9+7;vector<LL>v[maxn];map<LL,LL>M;LL dp[maxn][maxn],p[maxn],k;int n,m;LL gcd(LL a,LL b){    return b?gcd(b,a%b):a;}LL lcm(LL a,LL b){    return a/gcd(a,b)*b;}LL dfs(LL u,LL s){//    cout<<"madan  "<<u<<endl;    int num=M[s];    if(dp[u][num])   return dp[u][num];    LL ret=0;    for(int i=0;i<(int)v[u].size();i++)    {        LL ed=v[u][i];        LL LCM=lcm(s,p[ed]);        if(k%LCM||LCM==s&&u!=0)  continue;        ret+=dfs(ed,LCM)%MOD;    }    return dp[u][num]=ret;}int main(){    LL st,ed;    while(~scanf("%d%d%lld",&n,&m,&k))    {        REPF(i,0,n)  v[i].clear();        while(m--)        {            scanf("%lld%lld",&st,&ed);//wa了一个小时            v[st].push_back(ed);        }        REPF(i,1,n)   scanf("%lld",&p[i]);        if(k%p[n])        {            puts("0");            continue;        }        M.clear();        v[0].push_back(1);        int cnt=0;        for(LL i=1;i*i<=k;i++)        {            if(k%i==0)            {                M[i]=cnt++;                M[k/i]=cnt++;            }        }        memset(dp,0,sizeof(dp));        dp[n][M[k]]=1;        printf("%lld\n",dfs(0,1));    }    return 0;}