2015 Multi-University Training Contest 9-1007 Travelling Salesman Problem

最新推荐文章于 2024-07-10 16:09:18 发布

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最新推荐文章于 2024-07-10 16:09:18 发布

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原文链接：http://www.cnblogs.com/yutingliuyl/p/6849051.html

本文探讨了一种迷宫路径选择算法，旨在寻找从左上角到右下角的路径，使得路径上的数字之和最大。针对不同迷宫大小提出了有效的解决方案，并提供了完整的代码实现。

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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

这是一道有趣的问题，怎样才干从左上到右下使经过路径和最大，每一个数字都是正数，那么假设能走全，肯定是走全比較好，所以当n||m有奇数时。可直接构造之，假设均为偶数时，能够发现，我能够绕啊绕的绕过一个点，剩下的都遍历，横纵坐标和偶数的无法仅仅避开这一个点。所以要想绕开这个点，必需要附带至少一个其他点且是能够单独避开的奇点。所以我们仅仅要找到偶点中最小的那个点绕开就好了。

这样全部情况都构造出来了。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
void print(int x,int y,char dir)
{
    for(int i=x;i<y;i++)
        printf("%c",dir);
}

int f[111][111];
long long ret=0;
int x,y;
int m,n,mi;
int main()
{
    while(scanf("%d%d",&m,&n)==2)
    {
        ret=0;
        mi=0x3f3f3f3f;
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&f[i][j]);
                ret+=f[i][j];
                if((i+j)%2==1 && f[i][j]<mi)
                {
                    mi=f[i][j];
                    x=i;
                    y=j;
                } 
            }
        if(m%2 == 1|| n%2==1)
        {
            cout<<ret<<'\n';
            if(m%2==1)
            {
                for(int i=1;i<=m;i++)
                {
                    if(i%2==1)
                        print(1,n,'R');
                    else
                        print(1,n,'L');
                    if(i!=m)
                    {
                        print(i,i+1,'D');
                    }
                }
            }
            else
            {
                for(int i=1;i<=n;i++)
                {
                    if(i%2==1)
                        print(1,m,'D');
                    else
                        print(1,m,'U');
                    if(i!=n)
                        print(i,i+1,'R');
                }
            }
        }
        else
        {
            cout<<ret-f[x][y]<<'\n';
            if(x%2==1)
            {
                for(int i=1;i<x;i++)
                {
                    if(i%2==1)
                        print(1,n,'R');
                    else
                        print(1,n,'L');
                    print(i,i+1,'D');
                }
                for(int i=1;i<=n;i++)
                {
                    if(i<y)
                        if(i%2==1)
                            print(1,2,'D');
                        else
                            print(1,2,'U');
                    else
                        if(i>y)
                        {
                            if(i%2==0)
                                print(1,2,'D');
                            else
                                print(1,2,'U');
                        }
                    if(i!=n)
                        print(1,2,'R');
                }
                for(int i=x+2;i<=m;i++)
                {
                    print(1,2,'D');
                    if(i%2==1)
                        print(1,n,'L');
                    else
                        print(1,n,'R');
                }
            }
            else
            {
                for(int i=1;i<y;i++)
                {
                    if(i%2==1)
                        print(1,m,'D');
                    else
                        print(1,m,'U');
                    print(1,2,'R');
                }
                for(int i=1;i<=m;i++)
                {
                    if(i<x)
                    {
                        if(i%2==1)
                            print(1,2,'R');
                        else
                            print(1,2,'L');
                    }
                    else
                        if(i>x)
                        {
                            if(i%2==1)
                                print(1,2,'L');
                            else
                                print(1,2,'R');
                        }
                    if(i!=m)
                        print(1,2,'D');
                }
                for(int i=y+2;i<=n;i++)
                {
                    print(1,2,'R');
                    if(i%2==1)
                        print(1,m,'U');
                    else
                        print(1,m,'D');
                }
            }
        }
        printf("\n");
    }
    return 0;
}

转载于:https://www.cnblogs.com/yutingliuyl/p/6849051.html