hdu 5402 ，Travelling Salesman Problem,2015多校联合训练赛#9

Travelling Salesman Problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 183    Accepted Submission(s): 70
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
   
   
   
   

 

Author

xudyh

 

Source

2015 Multi-University Training Contest 9

当n,m有奇数时，可以选择一条蛇形道路全部点都取到。算总和就行。

如果n，m为偶数时，选择(i+j)%==1的位置中最小的点不走，其他全部点选择即可。

还是蛇形道路，但是对于选择不走的那个位置所在行和相邻行，要特殊处理，优先上下方向走就行。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int num[111][111];

int n,m;
void work(){
    long long ans = 0;
    for(int i = 0;i < n; i++)
        for(int j = 0;j < m; j++)
            ans += num[i][j];
    printf("%I64d\n",ans);
    if(n % 2 == 1){
        for(int i = 0;i < n; i++){
            for(int j = 1;j < m; j++){
                if(i%2==0) putchar('R');
                else putchar('L');
            }
            if(i != n - 1) putchar('D');
        }
    }
    else {
        for(int i = 0;i < m; i++){
            for(int j = 1;j < n; j++){
                if(i%2  == 0) putchar('D');
                else putchar('U');
            }
            if(i != m-1)putchar('R');
        }
    }
}
int check[111][111];
int ok(int x,int y){
    if(x < 0 || x == n || y < 0 || y == m) return 0;
    if(check[x][y] == 1) return 0;
    return 1;
}
void dfs(int x,int y,int c1,int c2){
    check[x][y] = 1;
    if(x == c1){
        if(ok(x+1,y)){
            putchar('D');
            dfs(x+1,y,c1,c2);
        }
        else if(ok(x,y+1)){
            putchar('R');
            dfs(x,y+1,c1,c2);
        }
        else if(ok(x,y-1)){
            putchar('L');
            dfs(x,y-1,c1,c2);
        }
    }
    else if(x == c2){
        if(ok(x-1,y)){
            putchar('U');
            dfs(x-1,y,c1,c2);
        }
        else if(ok(x,y+1)){
            putchar('R');
            dfs(x,y+1,c1,c2);
        }
        else if(ok(x,y-1)){
            putchar('L');
            dfs(x,y-1,c1,c2);
        }
        else if(ok(x+1,y)){
            putchar('D');
            dfs(x+1,y,c1,c2);
        }
    }
    else {
        if(ok(x,y+1)){
            putchar('R');
            dfs(x,y+1,c1,c2);
        }
        else if(ok(x,y-1)){
            putchar('L');
            dfs(x,y-1,c1,c2);
        }
        else if(ok(x+1,y)){
            putchar('D');
            dfs(x+1,y,c1,c2);
        }
    }
}
void work2(){
    long long ans = 0, res = 10000000000ll;
    int c1,c2,y;
    for(int i = 0 ;i < n; i++)
        for(int j = 0;j < m; j++){
            ans += num[i][j];
            if((i+j)% 2 == 1){
                if(res > num[i][j]){
                    res = num[i][j];
                    c1 = i;
                    y = j;
                }
            }
        }
    printf("%I64d\n",ans-res);
    if(c1 == 0) c2 = 1;
    else c2 = c1-1;
    memset(check,0,sizeof(check));
    check[c1][y] = 1;
    dfs(0,0,min(c1,c2),max(c1,c2));
}
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i = 0;i < n; i++)
            for(int j = 0;j < m; j++)
                scanf("%d",&num[i][j]);
        if( n%2 == 1 || m % 2 == 1) work();
        else work2();
        printf("\n");
    }
    return 0;
}

/*

3 3
2 3 3
3 3 3
3 3 2
3 2
2 3
3 3
3 3
2 3
2 3 3
3 3 3
4 4
1 2 3 4
5 6 7 8
9 8 7 6
5 4 3 2

2 2
1 2
5 2

1 2
1 2

2 1
2 1

*/