HDU 5402 Travelling Salesman Problem(棋盘黑白染色)——多校练习9

Travelling Salesman Problem

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

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

 
/*********************************************************************/

题意：现有一个n*m的迷宫(n行m列)，每一个格子都有一个非负整数，Teacher Mai想要从迷宫的左上角(1,1)到迷宫的右下角(n,m)，并且使得他走过的路径的整数之和最大，问最大和为多少以及他走的路径。
放上出题人的解题报告



解题思路：首先，因为每个格子都是非负整数，而且规定每个格子只能走一次，所以为了使和尽可能大，必定是走的格子数越多越好。这样我们就需要考虑一下是不是所有的格子都可以走。

在纸上画画，你就会发现，若n、m中至少有一个是奇数的话，必然能够遍历每一个格子，这样的话，我们只需往n、m中为偶数的那个方向先走就可以了





若n、m都为偶数的话，根据棋盘黑白染色(关于棋盘黑白染色问题，想了解的可以点链接)可以得知，当假设(1,1)与(n,m)都为黑色，那么这条路径势必黑色格子数会比白色格子数多1，而棋盘中黑白格子数是相等的，所以棋盘中有一个白格子不会被经过

或许你自己在研究这道题的时候，会感觉有点混乱，总想着删值最小的格子，但有些格子删了，会有好几个格子走不到，那是因为删了黑格子的缘故，那样导致黑白格子数差2，又要有两个白格子无法到达，这样和势必会比只删一个白格子要来得小。



所以只能删白格子



其他的在此不再细讲，可以看出题人的解题报告，如果需要我对一些细节进行补充的话，也可以提出来

暂时先放上几组数据以供参考

3 4

2 3 3 3

3 3 3 3

3 3 3 3

35

RRRDLLLDRRR

4 3

2 3 3

3 3 3

3 3 3

3 3 3

35

DDDRUUURDDD

4 4

3 2 3 3

3 3 3 3

3 3 3 3

3 3 3 3

45

DRRURDDLLLDRRR

4 4

3 3 3 3

3 3 3 3

3 2 3 3

3 3 3 3

45

RRRDLLLDDRRURD

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 105;
const int inf = 1000000000;
const int mod = 1000000007;
int s

;
int main()
{
int n,m,i,j,sum,x,y;
while(~scanf("%d%d",&n,&m))
{
sum=0;x=1,y=2;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
scanf("%d",&s[i][j]);
sum+=s[i][j];
if(((i+j)&1)&&s[x][y]>s[i][j])
x=i,y=j;
}
if(n&1||m&1)
{
printf("%d\n",sum);
if(n&1)
{
for(i=1;i<=n;i++)
{
for(j=1;j<m;j++)
if(i&1)
printf("R");
else
printf("L");
if(i<n)
printf("D");
else
puts("");
}
}
else
{
for(i=1;i<=m;i++)
{
for(j=1;j<n;j++)
if(i&1)
printf("D");
else
printf("U");
if(i<m)
printf("R");
else
puts("");
}
}
}
else
{
printf("%d\n",sum-s[x][y]);
for(i=1;i<=n;i+=2)
{
if(x==i||x==i+1)
{
for(j=1;j<y;j++)
{
if(j&1)
printf("D");
else
printf("U");
printf("R");
}
if(y<m)
printf("R");
for(j=y+1;j<=m;j++)
{
if(j&1)
printf("U");
else
printf("D");
if(j<m)
printf("R");
}
if(i<n-1)
printf("D");
}
else if(x<i)
{
for(j=1;j<m;j++)
printf("L");
printf("D");
for(j=1;j<m;j++)
printf("R");
if(i<n-1)
printf("D");
}
else
{
for(j=1;j<m;j++)
printf("R");
printf("D");
for(j=1;j<m;j++)
printf("L");
printf("D");
}
}
puts("");
}
}
return 0;
}

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