HDU 5607 求A到B走K步的概率 矩阵快速幂DP

#include<cstdio>
#include<vector>
#define LL long long
using namespace std;
const int N = 55;
const int MO = 1e9+7;
int n,m,u,k,T,out
;
struct Mat{
int h,l;
LL f

;
void clear(){
for(int i=1;i<=h;++i) for(int j=1;j<=l;++j) f[i][j]=0;
}
Mat(int _h,int _l){ h=_h; l=_l; clear(); }
Mat friend unit(Mat x){
Mat ret=Mat(x.h,x.l);
for(int i=0;i<=x.h;++i) ret.f[i][i]=1;
return ret;
}
Mat friend operator * (Mat A,Mat B){
Mat ret=Mat(A.h,B.l);
for(int i=0;i<=ret.h;++i){
for(int j=0;j<=ret.l;++j){
for(int k=0;k<=A.l;++k){
ret.f[i][j]+=A.f[i][k]*B.f[k][j];
ret.f[i][j]%=MO;
}
}
}
return ret;
}
};
Mat qpow(Mat x,int k){
Mat ret=unit(x),t=x;
for(;k;k>>=1){
if(k&1)ret=ret*t; t=t*t;
}
return ret;
}
/*LL qpow(int x,int k){
LL ret=1;
for(;k;k>>=1){ if(k&1) (ret*=x)%=MO; (x*=x)%=MO; }
return ret;
}*/
LL exgcd(LL a,LL b,LL &x,LL &y){
if(!b) return x=1,y=0,a;
LL r=exgcd(b,a%b,x,y),t=x;
return x=y,y=t-a/b*y,r;
}
LL Ine(LL x){
LL Y,rev;
exgcd(x,MO,rev,Y);
rev = (rev%MO+MO)%MO;
return rev;
}

int main(){
//freopen("graph.in","r",stdin);
scanf("%d%d",&n,&m);
Mat M=Mat(n,n);
for(int x,y,i=1;i<=m;++i){
scanf("%d%d",&x,&y);
++M.f[x][y]; ++out[x];
}
for(int i=1;i<=n;++i){
LL ni=Ine(out[i]);
for(int j=1;j<=n;++j){
(M.f[i][j]*=ni)%MO;
}
}
scanf("%d",&T);
while(T--){
scanf("%d%d",&u,&k);
Mat Ans=Mat(n,n);
Ans.f[u][u]=1;
Mat tM=qpow(M,k);
Ans=Ans*tM;
for(int i=1;i<=n;++i){
printf("%I64d ",Ans.f[u][i]);
}
printf("\n");
}
return 0;
}