Coin-collecting by robot

最新推荐文章于 2023-02-03 18:01:05 发布

CGD_ProgramLife

最新推荐文章于 2023-02-03 18:01:05 发布

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分类专栏： c/c++ 算法分析文章标签： Coin-collecting by robot 算法分析

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本文探讨了机器人进行硬币收集的问题，详细分析了输入、输出及其样例，揭示了背后的算法逻辑。

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Coin-collecting by robot

1000(ms)

65535(kb)

738 / 2013

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

#include<iostream> 
#include<stdio.h> 
#include<malloc.h>
#include<string.h>
#include<string>
#include<stack>
#include<math.h>
#include<algorithm>
using namespace std;
#define INF 99999999

int a[1005][1005];

int main() {
	int m,n;
	cin>>m>>n;
	for(int i=1;i<=m;i++){
		for(int j=1;j<=n;j++){
			cin>>a[i][j];
		}
	}
	for(int i=1;i<=m;i++){
		for(int j=1;j<=n;j++){
			if(a[i-1][j]>a[i][j-1])
				a[i][j]+=a[i-1][j];
			else
				a[i][j]+=a[i][j-1];
		}
	}
	cout<<a[m][n]<<endl;
 	return 0;
}