ZOJ 3644 Kitty's Game (广搜，STL中的map,vector)

Kitty's Game

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 84

1 2

2 5

1 3

3 5

1 4

4 5

1 5 4 12 21

Sample Output

2




此题比赛的时候没有想到方法。
赛后看题解恍然大悟。
用map映射可以很方便解决。
直接搜索就可以了。
具体看代码：

#include<stdio.h>

#include<vector>

#include<iostream>

#include<map>

#include<string.h>

#include<queue>

#include<algorithm>

using namespace std;



const int MAXN=2010;

const int MOD=1000000007;

vector<int>g[MAXN];

map<int,int>dp[MAXN];

int n,k;

int score[MAXN];

bool vis[MAXN];



int gcd(int a,int b)

{

    if(b==0)return a;

    return gcd(b,a%b);

}

int bfs()

{

    int ret=0;

    queue<int>q;

    dp[1][score[1]]=1;

    memset(vis,false,sizeof(vis));

    while(!q.empty())q.pop();

    q.push(1);

    vis[1]=true;

    while(!q.empty())

    {

        int cur=q.front();

        q.pop();

        vis[cur]=false;

        if(cur==n)ret=(ret+dp[cur][k])%MOD;

        for(int i=0;i<g[cur].size();i++)

        {

            int v=g[cur][i];

            map<int,int>::iterator it;

            for(it=dp[cur].begin();it!=dp[cur].end();it++)

            {

                int temp=it->first;

                int lcm=temp/gcd(temp,score[v])*score[v];

                if(lcm!=temp &&  k%lcm==0)

                {

                    dp[v][lcm]=(dp[v][lcm]+it->second)%MOD;

                    if(!vis[v])

                    {

                        vis[v]=true;

                        q.push(v);

                    }

                }

            }

        }

        dp[cur].clear();

    }

    return ret;

}

int main()

{

    int m,u,v;

    while(scanf("%d%d%d",&n,&m,&k)!=EOF)

    {

        for(int i=1;i<=n;i++)g[i].clear();

        for(int i=1;i<=n;i++)dp[i].clear();

        while(m--)

        {

            scanf("%d%d",&u,&v);

            g[u].push_back(v);

        }

        for(int i=1;i<=n;i++)scanf("%d",&score[i]);

        printf("%d\n",bfs());

    }

    return 0;

}