hdu2157 how many ways 矩阵

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邻接矩阵自乘k次中对应的即为方案数

因为乘法原则，所以每次乘都会统计出来

这想法不错

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<algorithm>#define fo(i,a,b) for(int i=a;i<=b;i++)#define down(i,a,b) for(int i=a;i>=b;i--)using namespace std;#define N 35#define inf 1000int n,m;struct matrix{int a[N][N];void clear(){memset(a,0,sizeof(a));}}I;matrix operator*(const matrix a,const matrix b){matrix anss;fo(i,0,m-1)fo(j,0,m-1){anss.a[i][j]=0;fo(k,0,m-1){anss.a[i][j]+=a.a[i][k]*b.a[k][j];anss.a[i][j]%=inf;}}return anss;}void getI(){I.clear();fo(i,0,m-1)fo(j,0,m-1)if(i==j)I.a[i][j]=1;else I.a[i][j]=0;}matrix KSM(matrix a,int k){matrix ret=I;while(k){if(k&1)ret=a*ret;a=a*a;k>>=1;}return ret;}int main(){matrix A,ans;while(scanf("%d%d",&m,&n)!=EOF&&(n||m)){getI();A.clear();fo(i,1,n){int x,y;scanf("%d%d",&x,&y);A.a[x][y]=1;}int t;scanf("%d",&t);while(t--){int x,y,k;scanf("%d%d%d",&x,&y,&k);ans=KSM(A,k);printf("%d\n",ans.a[x][y]);}}return 0;}

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