hdu5402 Travelling Salesman Problem

最新推荐文章于 2024-07-21 14:30:00 发布

kingsirre

最新推荐文章于 2024-07-21 14:30:00 发布

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分类专栏： 2015 Multi-University Training

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本文深入探讨了深度学习在人工智能领域的应用，包括卷积神经网络、循环神经网络、强化学习等关键技术，以及它们在图像处理、自然语言处理、语音识别、推荐系统等多个领域的实际应用案例。

Problem Description

Teacher Mai is in a maze with

n rows and

m columns. There is a non-negative number in each cell. Teacher Mai wants to walk from the top left corner

(1,1) to the bottom right corner

(n,m). He can choose one direction and walk to this adjacent cell. However, he can't go out of the maze, and he can't visit a cell more than once.

Teacher Mai wants to maximize the sum of numbers in his path. And you need to print this path.

Input

There are multiple test cases.

For each test case, the first line contains two numbers

n,m(1≤n,m≤100,n∗m≥2).

In following

n lines, each line contains

m numbers. The

j-th number in the

i-th line means the number in the cell

(i,j). Every number in the cell is not more than

104.

Output

For each test case, in the first line, you should print the maximum sum.

In the next line you should print a string consisting of "L","R","U" and "D", which represents the path you find. If you are in the cell

(x,y), "L" means you walk to cell

(x,y−1), "R" means you walk to cell

(x,y+1), "U" means you walk to cell

(x−1,y), "D" means you walk to cell

(x+1,y).

Sample Input

Sample Output

25
RRDLLDRR

这道题就是找规律暴力，也没有简化成什么。注意有的情况不是可以全部遍历到。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>

using namespace std;

#define MAXN 100010
int main()
{
    int n,m,a;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int sum=0,minn=MAXN,x=0,y=0;
       for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        {
            scanf("%d",&a);
            sum+=a;
            if(minn>a&&(i+j)%2)
            {
                x=i;
                y=j;
                minn=a;
            }
        }
        if(n%2==1)
        {
            printf("%d\n",sum);
            for(int i=0;i<n/2;i++)
            {
                for(int j=0;j<m-1;j++)
                    printf("R");
                printf("D");
                for(int j=0;j<m-1;j++)
                printf("L");
                printf("D");
            }
            for(int j=0;j<m-1;j++)
                printf("R");
        }
        else if(m%2==1)
        {
            printf("%d\n",sum);
            for(int i=0;i<m/2;i++)
            {
                for(int j=0;j<n-1;j++)
                printf("D");
                printf("R");
                for(int j=0;j<n-1;j++)
                printf("U");
                printf("R");
            }
            for(int j=0;j<n-1;j++)
                printf("D");
        }
        else
        {
            printf("%d\n",sum - minn);
            for(int i=0;i<x/2;i++)
            {
               for(int j=0;j<m-1;j++)
                    printf("R");
                printf("D");
                for(int j=0;j<m-1;j++)
                     printf("L");
                printf("D");
            }
            for(int j=0;j<y/2;j++)
                printf("DRUR");
            if(x%2==1)
                printf("RD");
            else
               printf("DR");
            for(int j=y/2+1;j<m/2;j++)
                printf("RURD");
            for(int i=x/2+1;i<n/2;i++)
            {
                 printf("D");
                 for(int j=0;j<m-1;j++)
                     printf("L");
                printf("D");
                for(int j=0;j<m-1;j++)
                    printf("R");
            }
        }
        puts("");
    }
    return 0;
}