2015 多校联赛 ——HDU5402（模拟）

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

3 3 2 3 3 3 3 3 3 3 2

 

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
",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
",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("
");
    }
    return 0;
}