zoj2271 Chance to Encounter a Girl(DP)

/*
概率计算：按时间为阶段，每个点由上一阶段周围的四个点来维护。
注意事项：1.时间O（N^3*T），在问题的边缘时间，所以打表计算。
　　　 2.关于概率的求解，如果遇到就结束了，所以向后走就说明
　　　　之前没有碰到，所以不用前面的碰到的概率计算后面的值。
*/

View Code

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 double answ[ 51 ] = {
 5     0.0000,0.6667,0.0000,0.4074,0.0000,
 6     0.3361,0.0000,0.2928,0.0000,0.2629,
 7     0.0000,0.2407,0.0000,0.2233,0.0000,
 8     0.2092,0.0000,0.1975,0.0000,0.1875,
 9     0.0000,0.1789,0.0000,0.1714,0.0000,
10     0.1648,0.0000,0.1589,0.0000,0.1536,
11     0.0000,0.1487,0.0000,0.1443,0.0000,
12     0.1403,0.0000,0.1366,0.0000,0.1332,
13     0.0000,0.1300,0.0000,0.1270,0.0000,
14     0.1243,0.0000,0.1217,0.0000,0.1192};
15 
16 /*
17 
18 double maps[ 100 ][ 100 ][ 100 ];
19 short  dxdy[ 4 ][ 2 ] = {1,0,0,1,-1,0,0,-1};
20 
21 void madelist()    // 打表程序 
22 {
23     for ( int n = 1 ; n < 100 ; n += 2 ) {
24         double sum = 0.0;
25         memset( maps, 0, sizeof( maps ) );
26         maps[ 0 ][ n/2 ][ n/2 ] = 1.0;
27         for ( int t = 0 ; t < n ; ++ t )
28         for ( int i = 0 ; i < n ; ++ i )
29         for ( int j = 0 ; j < n ; ++ j )
30         for ( int k = 0 ; k < 4 ; ++ k ) {
31             int x = i+dxdy[ k ][ 0 ];
32             int y = j+dxdy[ k ][ 1 ];
33             if ( x > 0 && x < n-1 && y > 0 && y < n-1 )
34                 maps[ t+1 ][ i ][ j ] += 0.25*maps[ t ][ x ][ y ];
35             else if ( ( x == 0 && y == 0 ) || ( x == 0 && y == n-1 ) ||
36                     ( x== n-1 && y == 0 ) || ( x == n-1 && y == n-1 ) )
37                 maps[ t+1 ][ i ][ j ] += 0.5*maps[ t ][ x ][ y ];
38             else maps[ t+1 ][ i ][ j ] += 1.0/3*maps[ t ][ x ][ y ];
39             if ( i == n/2 && j == t ) {
40                 sum += maps[ t+1 ][ i ][ j ];
41                 maps[ t+1 ][ i ][ j ] = 0.0;
42             }
43         }
44         printf("%.4lf,",sum);
45     }
46 }
47 */
48 
49 int main()
50 {
51     int m;
52     while ( ~scanf("%d",&m) ) 
53         printf("%.4lf\n",answ[ m/2 ]);
54     return 0;
55 }