swustoj1132Coin-collecting by robot

最新推荐文章于 2023-02-03 18:01:05 发布

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最新推荐文章于 2023-02-03 18:01:05 发布

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分类专栏： swustoj DP动态规划

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本文介绍了一个使用动态规划解决机器人在棋盘上收集硬币的问题。通过定义状态转移方程，求解机器人从左上角到右下角能收集到的最大硬币数量。

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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

思路：动态规划。

dp【i】【j】表示走到（i，j）这个点所能获得的最大coin数。

转移方程为：

mp[i][j] = max(mp[i - 1][j], mp[i][j - 1]);

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<queue>
#include<iostream>
#include<string.h>
using namespace std;
int mp[1005][1005];
//oj1132
int main()
{
	int n, m;
	while (cin >> n >> m) {
		for (int i = 0;i < n;i++) {
			for (int j = 0;j < m;j++) {
				scanf("%d", &mp[i][j]);
			}
		}
		for (int i = 1;i < n;i++)
		 {
			if (mp[i][0] == 1) {
				mp[i][0] = mp[i - 1][0]+1;
			}
			else {
				mp[i][0] = mp[i - 1][0];
			}
		}
		 for (int i = 1;i < m;i++) {
			 if (mp[0][i] == 1) {
				 mp[0][i] = mp[0][i - 1] + 1;
			 }
			 else {
				 mp[0][i] = mp[0][i - 1];
			 }
		 }
		for (int i = 1;i < n;i++) {
			for (int j = 1;j < m;j++) {
				if(mp[i][j])
				mp[i][j] = max(mp[i - 1][j], mp[i][j - 1])+1;
				else {
					mp[i][j] = max(mp[i - 1][j], mp[i][j - 1]);
				}
			}
		}
		int ans = mp[n - 1][m - 1];
		cout <<  ans <<"\r\n";
	}
	return 0;
}