zoj 2271 - Chance to Encounter a Girl

题目：平面图上有一个女孩，她初始在（n/2,，n/2），每次可以走到上下左右四个格子中的一个，

            她每次随机的走的动，你从（-1，n/2）向右移动，问你们相遇的概率。

分析：概率dp。事件为阶段，每个点由上一阶段周围的四个点来维护。

            分成角、边、和中间三种计算概率（分别为1/5，1/3，1/4）；

            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；

            时间O（N^3*T），在问题的边缘时间，所以打表计算。

说明：（2011-09-19 09:31）。

#include <stdio.h>#include <string.h>double answ[ 51 ] = {    0.0000,0.6667,0.0000,0.4074,0.0000,    0.3361,0.0000,0.2928,0.0000,0.2629,    0.0000,0.2407,0.0000,0.2233,0.0000,    0.2092,0.0000,0.1975,0.0000,0.1875,    0.0000,0.1789,0.0000,0.1714,0.0000,    0.1648,0.0000,0.1589,0.0000,0.1536,    0.0000,0.1487,0.0000,0.1443,0.0000,    0.1403,0.0000,0.1366,0.0000,0.1332,    0.0000,0.1300,0.0000,0.1270,0.0000,    0.1243,0.0000,0.1217,0.0000,0.1192};/*double maps[ 100 ][ 100 ][ 100 ];short  dxdy[ 4 ][ 2 ] = {1,0,0,1,-1,0,0,-1};void madelist()    // 打表程序 {    for ( int n = 1 ; n < 100 ; n += 2 ) {        double sum = 0.0;        memset( maps, 0, sizeof( maps ) );        maps[ 0 ][ n/2 ][ n/2 ] = 1.0;        for ( int t = 0 ; t < n ; ++ t )        for ( int i = 0 ; i < n ; ++ i )        for ( int j = 0 ; j < n ; ++ j )        for ( int k = 0 ; k < 4 ; ++ k ) {            int x = i+dxdy[ k ][ 0 ];            int y = j+dxdy[ k ][ 1 ];            if ( x > 0 && x < n-1 && y > 0 && y < n-1 )                maps[ t+1 ][ i ][ j ] += 0.25*maps[ t ][ x ][ y ];            else if ( ( x == 0 && y == 0 ) || ( x == 0 && y == n-1 ) ||                    ( x== n-1 && y == 0 ) || ( x == n-1 && y == n-1 ) )                maps[ t+1 ][ i ][ j ] += 0.5*maps[ t ][ x ][ y ];            else maps[ t+1 ][ i ][ j ] += 1.0/3*maps[ t ][ x ][ y ];            if ( i == n/2 && j == t ) {                sum += maps[ t+1 ][ i ][ j ];                maps[ t+1 ][ i ][ j ] = 0.0;            }        }        printf("%.4lf,",sum);    }}*/int main(){    int m;    while ( ~scanf("%d",&m) )         printf("%.4lf\n",answ[ m/2 ]);    return 0;}