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  • cf 429 B Working out

    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.

    Examples
    input
    Copy
    3 3
    100 100 100
    100 1 100
    100 100 100
    output
    Copy
    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].

    题目意思:

    给n*m的矩阵,每个格子有个数,A从(1,1)出发只能向下或右走,终点为(n,m),B从(n,1)出发只能向上或右走,终点为(1,m)。两个人的速度不一样,走到的格子可以获的该格子的数,两人相遇的格子上的数两个人都不能拿。求A和B能拿到的数的总和的最大值。

    n,m<=1000

    解题思路:

    dp.

    先预处理出每个格子到四个角落格子的路径最大数值,然后枚举两个人相遇的交点格子,枚举A、B的进来和出去方式,求最大值即可。

    注意边界情况。

    #include<iostream>
    #include<string.h>
    #include<string>
    #include<cmath>
    using namespace std;
    
    typedef long long ll;
    const int N = 1e3+10;
    int n,m;
    ll f1[N][N], f2[N][N], f3[N][N], f4[N][N];
    ll a[N][N];
    
    ll getRes(int i,int j) {
        return max( f1[i][j-1] + f4[i][j+1] + f2[i+1][j] + f3[i-1][j]
                , f1[i-1][j] + f4[i+1][j] + f2[i][j-1] + f3[i][j+1]);
    }
    
    
    int main () {
        //freopen("./Desktop/in.txt","r",stdin);
        scanf("%d %d", &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++)
                f1[i][j] = max(f1[i-1][j], f1[i][j-1]) + a[i][j];
        for(int i=n;i>=1;i--) 
            for(int j=1;j<=m;j++)
                f2[i][j] = max(f2[i][j-1], f2[i+1][j]) + a[i][j];
        for(int i=1;i<=n;i++) 
            for(int j=m;j>=1;j--)
                f3[i][j] = max(f3[i-1][j], f3[i][j+1]) + a[i][j];
        for(int i=n;i>=1;i--) 
            for(int j=m;j>=1;j--)
                f4[i][j] = max(f4[i+1][j], f4[i][j+1]) + a[i][j];
        ll mx=0;
        for(int i=2;i<n;i++) {
            for(int j=2;j<m;j++) {
                ll res = getRes(i,j);
                mx = max(mx, res);
                //printf("%lld ",res);
            }//puts("");
        }
        cout << mx <<endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Draymonder/p/9506867.html
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