题目描述

Several coins are placed in cells of an n×m board. A robot, located in the upper left cell of the board, needs to collect as many of the coins as possible and bring them to the bottom right cell. On each step, the robot can move either one cell to the right or one cell down from its current location.

输入

The fist line is n,m, which 1< = n,m <= 1000.
Then, have n row and m col, which has a coin in cell, the cell number is 1, otherwise is 0.

输出

The max number Coin-collecting by robot.

样例输入

5 6
0 0 0 0 1 0
0 1 0 1 0 0
0 0 0 1 0 1
0 0 1 0 0 1
1 0 0 0 1 0

样例输出

5

代码

#include<iostream>
using namespace std;
int Coor[1000][1000];
int main(){
	int N, M;
	cin>>N>>M;
	for(int i = 1; i <= N; i++){
		for(int j = 1; j <= M; j++){
			cin>>Coor[i][j];
			if(i == 1 && j > 1) Coor[1][j] += Coor[1][j-1];
			if(j == 1 && i > 1) Coor[i][1] += Coor[i-1][1];
			if(i > 1 && j > 1) Coor[i][j] += max(Coor[i][j-1],Coor[i-1][j]);
		}
	}  
	cout<<Coor[N][M]<<endl;
	return 0;
}

思路

1. 动态规划问题；

2. 令F( i , j )为机器人截止到第 i 行第 j 列单元格 ( i , j ) 能够收集到的最大硬币数, 有如下递推式：
   

3. 值得注意的是：递归算法解题相对常用的算法如普通循环等，运行效率较低。因此，应该尽量避免使用递归，除非没有更好的算法或者某种特定情况，递归更为适合的时候。在递归调用的过程当中系统为每一层的返回点、局部量等开辟了栈来存储。递归次数过多容易造成栈溢出等。

4. 我第一次用了递归的方式，出现了 Time Limit Exceeded ,普通循环修正以后，便可以通过。

5. 但可以 利用上面的公式，我们可以通过逐行或者逐列的方式填充 n 乘以 m 表以求得F( i , j )。