2015 多校联赛 ——HDU5402（模拟）-优快云博客

本文链接：https://blog.youkuaiyun.com/Fun_Zero/article/details/47776373

此博客讨论了如何在给定的二维矩阵中找到从左上角到右下角的最大路径和，并提供了相应的行走路径表示。对于矩阵大小为偶数的情况，特别考虑了在特定点绕行以达到最大路径和。

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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

要求从左上角走到右下角的最大值。

如果n，m中有奇数则可以全部走完。否则需要在（i+j-2）%2 == 1的点中选择一个最小值绕过。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MAXN 300005
#define MIN 0
#define MAX 1000001


int main()
{
    int n,m;
    ll sum;
    char ch;
    int x;
    //freopen("1007.txt","r",stdin);
    while(scanf("%d%d",&n,&m) != EOF)
    {
        sum = 0;
        int minx = 100000;
        int mini=-1,minj=-1;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
            {
                scanf("%d",&x);
                sum += x;
                if((i+j-2)%2)
                {
                    if(x < minx)
                    {
                        minx = x;
                        mini = i;
                        minj = j;
                    }
                }
            }
        if(n % 2 || m % 2)
        {
            printf("%I64d\n",sum);
            if(n % 2)
            {
                for(int k = 1; k <= n; k++)
                {
                    if(k % 2)
                        ch = 'R';
                    else
                        ch = 'L';
                    for(int i = 1; i <= m-1; i++)
                        printf("%c",ch);
                    if(k != n)
                        printf("D");
                }

            }
            else
            {
                for(int k = 1; k <= m; k++)
                {
                    if(k % 2)
                        ch = 'D';
                    else
                        ch = 'U';
                    for(int i = 1; i <= n-1; i++)
                        printf("%c",ch);
                    if(k != m)
                        printf("R");
                }
            }
        }
        else
        {
            printf("%I64d\n",sum-minx);
            int k=1;
            while(1)
            {
                if(mini%2&&k==mini) break;
                if(mini%2==0&&(k+1)==mini) break;
                for(int i=2; i<=m; i++)
                    if(k%2) printf("R");
                    else printf("L");
                printf("D");
                k++;
            }
            if(mini%2)
            {
                int cx=k;
                int cy=1;
                while((cy+1)!=minj)
                {
                    printf("DRUR");
                    cy+=2;
                }
                printf("DR");
                cx++;
                cy++;
                while(cy!=m)
                {
                    printf("RURD");
                    cy+=2;
                }
                k+=2;
            }
            else
            {
                int cx=k;
                int cy=1;
                while(cy!=minj)
                {
                    printf("DRUR");
                    cy+=2;
                }
                printf("RD");
                cx++;
                cy++;
                while(cy!=m)
                {
                    printf("RURD");
                    cy+=2;
                }
                k+=2;
            }
            for(int i=k; i<=n; i++)
            {
                printf("D");
                for(int j=2; j<=m; j++)
                    if(i%2) printf("L");
                    else printf("R");
            }
        }
        printf("\n");
    }
    return 0;
}