BestCoder Round #68 (div.1) B 矩阵乘法优化DP

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显然是DP。。f[i][j]表示当前在i点走了j步，那么方程：
f[i][j]=sigma(f[pre[i]][j-1]/ind[pre[i]]),ind[i]表示i点的出度，pre[i]表示指向i的点。。
可以发现每次转移都是一样的。直接矩阵转移。

#include<iostream>#include<cstdio>#include<cstring>#define p 1000000007#define ll long long using namespace std;int n,m,Q;ll ind[55],a[55][55];struct M{int x,y;ll m[55][55];} f,g,b;inline int read(){    int a=0,f=1; char c=getchar();    while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();}    while (c>='0'&&c<='9') {a=a*10+c-'0'; c=getchar();}    return a*f;}inline ll quick_power(ll a,ll b){    ll ans=1;    for (ll i=b;i;a=a*a%p,i>>=1)        if (i&1) ans=ans*a%p;    return ans;}inline M mul(M a,M b){    M c;    memset(c.m,0,sizeof(c.m));    c.x=a.x; c.y=b.y;    for (int i=1;i<=a.x;i++)        for (int j=1;j<=b.y;j++)            for (int k=1;k<=a.y;k++)                c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%p;    return c;}int main(){    n=read(); m=read();    for (int i=1;i<=m;i++)    {        int u=read(),v=read();        ind[u]++; a[u][v]++;    }    for (int i=1;i<=n;i++)    {        ind[i]=quick_power(ind[i],p-2);        for (int j=1;j<=n;j++)             if (a[i][j]) b.m[i][j]=ind[i];    }    b.x=b.y=n;    Q=read();    while (Q--)    {        int u=read(),k=read();        memset(f.m,0,sizeof(f.m));        f.m[1][u]=1; f.x=1; f.y=n;        g=b;        while (k)        {            if (k&1) f=mul(f,g);            g=mul(g,g);            k>>=1;        }        for (int i=1;i<=n;i++) printf("%lld ",f.m[1][i]);        puts("");    }    return 0;}

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