P1535 游荡的奶牛

题目描述

Searching for the very best grass, the cows are travelling about the pasture which is represented as a grid with N rows and M columns (2 <= N <= 100; 2 <= M <= 100). Keen observer Farmer John has recorded Bessie's position as (R1, C1) at a certain time and then as (R2, C2) exactly T (0 < T <= 15) seconds later. He's not sure if she passed through (R2, C2) before T seconds, but he knows she is there at time T.

FJ wants a program that uses this information to calculate an integer S that is the number of ways a cow can go from (R1, C1) to (R2, C2) exactly in T seconds. Every second, a cow can travel from any position to a vertically or horizontally neighboring position in the pasture each second (no resting for the cows). Of course, the pasture has trees through which no cow can travel.

Given a map with '.'s for open pasture space and '*' for trees, calculate the number of possible ways to travel from (R1, C1) to (R2, C2) in T seconds.

奶牛们在被划分成N行M列(2 <= N <= 100; 2 <= M <= 100)的草地上游走，试图找到整块草地中最美味的牧草。Farmer John在某个时刻看见贝茜在位置 (R1, C1)，恰好T (0 < T <= 15)秒后，FJ又在位置(R2, C2)与贝茜撞了正着。 FJ并不知道在这T秒内贝茜是否曾经到过(R2, C2)，他能确定的只是，现在贝茜在那里。设S为奶牛在T秒内从(R1, C1)走到(R2, C2)所能选择的路径总数，FJ希望有一个程序来帮他计算这个值。每一秒内，奶牛会水平或垂直地移动1单位距离（奶牛总是在移动，不会在某秒内停在它上一秒所在的点）。草地上的某些地方有树，自然，奶牛不能走到树所在的位置，也不会走出草地。现在你拿到了一张整块草地的地形图，其中'.'表示平坦的草地，'*'表示挡路的树。你的任务是计算出，一头在T秒内从(R1, C1)移动到(R2, C2)的奶牛可能经过的路径有哪些。

输入输出格式

输入格式：

第1 行: 3 个用空格隔开的整数：N，M，T 。第2..N+1 行: 第i+1 行为M 个连续的字符，描述了草地第i 行各点的情况，保证字符是'.'和'*'中的一个。第N+2 行: 4 个用空格隔开的整数：R1，C1，R2，C2 。

输出格式：

第1 行: 输出S，含义如题中所述。

输入输出样例

输入样例#1：

4 5 6

...*.

...*.

.....

.....

1 3 1 5

输出样例#1：

说明

样例说明:

草地被划分成4 行5 列，奶牛在6 秒内从第1 行第3 列走到了第1 行第5 列。

奶牛在6 秒内从(1,3)走到(1,5)的方法只有一种（绕过她面前的树）

居然可以用动规!!!

f[k][i][j] 表示第 k 秒到达 (i,j) 的方案

 #include<cstdio>

 #include<iostream>

 using namespace std;

 const int N = ;

 int f[][N][N];

 int mp[N][N];

 int dx[] = {,,,-};

 int dy[] = {,-,,};

 int n,m,t,sx,sy,ex,ey;

 char ch;

 int main()

 {

     scanf("%d%d%d",&n,&m,&t);

     for(int i=;i<=n;++i)

         for(int j=;j<=m;++j)

         {

             cin>>ch;

             if(ch=='.') mp[i][j] = ;

         }

     scanf("%d%d%d%d",&sx,&sy,&ex,&ey);

     f[][sx][sy] = ;

     for(int k=;k<=t;++k)

         for(int i=;i<=n;++i)

             for(int j=;j<=m;++j)

                 if(mp[i][j]) for(int h=;h<;++h)

                 {

                     int xx = i+dx[h];

                     int yy = j+dy[h];

                     if(xx> && yy> && xx<=n && yy<=m && mp[xx][yy])

                         f[k][i][j] += f[k-][xx][yy];

                 }

     printf("%d",f[t][ex][ey]);

     return ;

 }