hdu-5402（多校2015）

Travelling Salesman Problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 258    Accepted Submission(s): 95

Special Judge

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

3 3
2 3 3
3 3 3
3 3 2
        

Sample Output

25
RRDLLDRR
  
  

  
  
   題意：給一個n*m的正整數棋盤，求從（1，1）走到（n，m）路徑和最大，并輸出一組走法。
  
  
   思路：
  
    
   
        
  
           
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<stdio.h>
#include<math.h>
#include <string>
#include<string.h>
#include<map>
#include<queue>
#include<set>
#include<utility>
#include<vector>
#include<algorithm>
#include<stdlib.h>
using namespace std;
#define eps 1e-8
#define inf 0x3f3f3f3f
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define ll long long int
#define mod 1000000007
#define maxn 900006
#define maxm 500010
int f[101][101],n,m;
int main()
{
    while(~scanf("%d%d",&n,&m)){
            int kx=-1;
            int ky=-1;
            int sum=0;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
        {
            rd(f[i][j]);
            sum+=f[i][j];
            if((i+j-2)%2){
                if(kx==-1) kx=i,ky=j;
                else if(f[i][j]<f[kx][ky]) kx=i,ky=j;
            }
        }
        if(n%2||m%2){
            printf("%d\n",sum);
            if(n&1)
            for(int i=1;i<=n;i++)
            {
                if(i!=1) printf("D");
                for(int j=2;j<=m;j++)
                    if(i%2) printf("R");
                else printf("L");
            }
            else if(m&1)
                for(int i=1;i<=m;i++)
            {
                if(i!=1) printf("R");
                for(int j=2;j<=n;j++)
                    if(i%2) printf("D");
                else printf("U");
            }
            printf("\n");
            continue;
        }
        printf("%d\n",sum-f[kx][ky]);
        int k=1;
        while(1){
            if(kx%2&&k==kx) break;
        if(kx%2==0&&(k+1)==kx) break;
            for(int i=2;i<=m;i++)
                if(k%2) printf("R");
            else printf("L");
            printf("D");
            k++;
        }
        if(kx%2){
            int cx=k;
            int cy=1;
            while((cy+1)!=ky){
                printf("DRUR");
                cy+=2;
            }
            printf("DR");cx++;cy++;
            while(cy!=m){
                printf("RURD");cy+=2;
            }
                    //cx--;cy++;
                k+=2;
            }
        else{
            int cx=k;
            int cy=1;
            while(cy!=ky){
                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;
}