[日常摸鱼]HDU2157 How many ways??

hhh我又开始水题目了

题意：给一张有向图，多次询问一个点到另一个点刚好走$k$步的方案数取模，点数很小

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每个$a,b,k$的询问直接把邻接矩阵$map$自乘$k$次后$map[a][b]$就是答案了，别问我怎么证x

话说这个题的范围还可以大好多的…$k$这么小不用快速幂应该都行

#include<cstdio>
#include<cstring>
const int MOD=1000;
const int N=25;
struct matrix
{
    int s[N][N];
    matrix(){memset(s,0,sizeof(s));}
};
int n,m,T;
inline matrix mul(matrix a,matrix b)
{
    matrix res;
    for(register int i=0;i<n;i++)
        for(register int j=0;j<n;j++)
            for(register int k=0;k<n;k++)
                res.s[i][j]=(res.s[i][j]+(a.s[i][k]*b.s[k][j])%MOD)%MOD;
    return res;
}
inline matrix pow(matrix a,int b)
{
    matrix res;
    for(register int i=0;i<n;i++)res.s[i][i]=1;
    for(;b;b>>=1,a=mul(a,a))if(b&1)res=mul(res,a);
    return res;
}
int main()
{
    //freopen("input.in","r",stdin);
    while(scanf("%d%d",&n,&m)==2&&(n+m))
    {
        matrix map;
        for(register int i=1;i<=m;i++)
        {
            int u,v;scanf("%d%d",&u,&v);
            map.s[u][v]=1;
        }scanf("%d",&T);
        while(T--)
        {
            int u,v,k;
            scanf("%d%d%d",&u,&v,&k);
            matrix a;a=map;
            a=pow(a,k);printf("%d
",a.s[u][v]%MOD);
        }
    }
    return 0;
}