Codeforces 429B B. Working out（DP）

最新推荐文章于 2024-08-28 22:49:31 发布

原创最新推荐文章于 2024-08-28 22:49:31 发布 · 207 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#动态规划 #ACM

算法学习专栏收录该内容

30 篇文章

订阅专栏

本文介绍了一种基于四个顶点在矩阵中寻找最佳相遇点的路径规划算法，通过预计算四个顶点到所有点的最佳路径，然后枚举所有可能的相遇点来确定总收益最大的方案。该算法适用于如健身房锻炼路径规划等场景。

B. Working out
time limit per test 2 seconds
memory limit per test 256 megabytes
input standard input
output standard output
Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where they go is a matrix a with n lines and m columns. Let number a[i][j] represents the calories burned by performing workout at the cell of gym in the i-th line and the j-th column.

Iahub starts with workout located at line 1 and column 1. He needs to finish with workout a[n][m]. After finishing workout a[i][j], he can go to workout a[i + 1][j] or a[i][j + 1]. Similarly, Iahubina starts with workout a[n][1] and she needs to finish with workout a[1][m]. After finishing workout from cell a[i][j], she goes to either a[i][j + 1] or a[i - 1][j].

There is one additional condition for their training. They have to meet in exactly one cell of gym. At that cell, none of them will work out. They will talk about fast exponentiation (pretty odd small talk) and then both of them will move to the next workout.

If a workout was done by either Iahub or Iahubina, it counts as total gain. Please plan a workout for Iahub and Iahubina such as total gain to be as big as possible. Note, that Iahub and Iahubina can perform workouts with different speed, so the number of cells that they use to reach meet cell may differs.

Input
The first line of the input contains two integers n and m (3 ≤ n, m ≤ 1000). Each of the next n lines contains m integers: j-th number from i-th line denotes element a[i][j] (0 ≤ a[i][j] ≤ 105).

Output
The output contains a single number — the maximum total gain possible.

Sample test(s)
input
3 3
100 100 100
100 1 100
100 100 100
output
800
Note
Iahub will choose exercises a[1][1] → a[1][2] → a[2][2] → a[3][2] → a[3][3]. Iahubina will choose exercises a[3][1] → a[2][1] → a[2][2] → a[2][3] → a[1][3].

解题思路

这道题可以通过理解为四个顶点相遇点的最佳路径，因为相遇点只有一个，所以每个相遇点有两种情况。先获取四个顶点到达所有其他点的最佳路径，后枚举所有可能的相遇点，即除了外围的所有点，取最大值就可以获得结果。

#include<bits/stdc++.h>
using namespace std;

int dp1[1010][1010];
int dp2[1010][1010];
int dp3[1010][1010];
int dp4[1010][1010];
int a[1010][1010];

int main(){
    ios::sync_with_stdio(false);

    int n,m;
    cin >> n >> m;

    for(int i = 1;i <= n; ++i){
        for(int j = 1;j <= m; ++j){
            cin >> a[i][j];
        }
    }

    for(int i = 1;i <= n; ++i){
        for(int j = 1;j <= m; ++j){
            dp1[i][j] = max(dp1[i-1][j],dp1[i][j-1]) + a[i][j];
        }
    }

    for(int i = n;i > 0; --i){
        for(int j = m;j > 0; --j){
            dp2[i][j] = max(dp2[i+1][j],dp2[i][j+1]) + a[i][j];
        }
    }

    for(int i = n;i > 0; --i){
        for(int j = 1;j <= m; ++j){
            dp3[i][j] = max(dp3[i+1][j],dp3[i][j-1]) + a[i][j];
        }
    }

    for(int i = 1;i <= n; ++i){
        for(int j = m;j > 0; --j){
            dp4[i][j] = max(dp4[i-1][j],dp4[i][j+1]) + a[i][j];
        }
    }


    int ans = 0;
    for(int i = 2; i < n; ++i){
        for(int j = 2;j < m; ++j){
            ans = max(ans,dp1[i][j-1]+dp2[i][j+1]+dp3[i+1][j]+dp4[i-1][j]);
            ans = max(ans,dp1[i-1][j]+dp2[i+1][j]+dp3[i][j-1]+dp4[i][j+1]);
        }
    }

    cout << ans << endl;
    return 0;
}