ZOJ 3497 Mistwald 矩陣快速冪

思路：

先將二維的座標轉換成一維的座標，然後用矩陣存儲，然後進行pci路徑轉移。關鍵點就是不能從出口處轉移。

#include <string>
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 26;
int n,m;
struct Matrix {
    int arr[N][N];
    Matrix(){};
    Matrix operator *( const Matrix &a ) const {
        Matrix ret;
        for(int i = 1; i <= m * n; i++) {
            for(int j = 1; j <= m * n; j++) {
                ret.arr[i][j] = 0;
                for(int k = 1; k <= m * n; k++) {
                    if ( i==m*n || k==m*n ) continue;
                    ret.arr[i][j] |= (arr[i][k] * a.arr[k][j]) ;
                }
            }
        }
        return ret;
    }
};
Matrix ans;
Matrix item,t;

void pow_M( int x ) {
    while ( x ) {
        if ( x&1 )
            ans = ans * item;
        item= item * item;
        x /= 2;
    }
}
int main()
{
    int T;
    cin>>T;
    int x[4],y[4];
    while ( T-- ) {
        scanf("%d%d",&n,&m);
        memset( item.arr,0,sizeof item.arr );
        memset( ans.arr,0,sizeof ans.arr );

        for ( int i=1; i<=m*n; i++ )
            ans.arr[i][i] =1 ;

        for ( int i=1; i<=n; i++ ) {
            for ( int j=1; j <=m; j++ ) {
                scanf(" ((%d,%d),(%d,%d),(%d,%d),(%d,%d))",&x[0],&y[0],&x[1],&y[1],&x[2],&y[2],&x[3],&y[3]);
                for(int k=0;k<4;k++)
                    item.arr[ (i-1)*m + j ][ (x[k]-1)*m + y[k] ] = 1;
            }
        }
        memcpy( t.arr , item.arr,sizeof item.arr );
        int q;
        int time;
        cin>>q;
        while ( q-- ) {

            scanf("%d",&time);
            memcpy( item.arr , t.arr , sizeof item.arr );
            memset( ans.arr,0,sizeof ans.arr );
            for ( int i=1; i<=m*n; i++ )
                ans.arr[i][i] =1 ;

            pow_M( time );
            int cnt = 0;
            for ( int i=1; i<=m*n; i++ ) {
                if ( ans.arr[1][i] )
                    ++cnt;
            }
            if ( ans.arr[ 1 ][ n*m ] ) {
                if ( cnt==1 ) cout<<"True"<<endl ;
                else cout<<"Maybe"<<endl;
            } else {
                cout<<"False"<<endl;
            }
        }
        cout<<endl;
    }

    return 0;
}