HDU 5402(Travelling Salesman Problem-构造矩阵对角最长不相交路径)

Travelling Salesman Problem

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 898 Accepted Submission(s): 327
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

Sample Output

25
RRDLLDRR

Author

xudyh

Source

2015 Multi-University Training Contest 9

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wange2014 | We have carefully selected several similar problems for you: 5421 5420 5419 5418 5417

当n,m有一个奇数时，‘S形’可全取。

否则至少要少取一个，

假设少取(mx,my) ，当mx+my为偶数时，必定有一个与(mx,my)相邻的不能取

否则必能全取剩下的。（‘S形’+特判2行‘长城形’)

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,m;
ll a[MAXN][MAXN];
int main()
{
//	freopen("Travelling.in","r",stdin);
	
	while(cin>>n>>m) {
		ll sum=0,mi=INF;int mx,my;
		For(i,n) For(j,m) {
			scanf("%lld",&a[i][j]),sum+=a[i][j];
			if (mi>a[i][j]&&(i+j)%2==1)
				mi=min(mi,a[i][j]),mx=i,my=j;
		}
		if (n%2==0&&m%2==0)
		{
			cout<<sum-mi<<endl;	
			
			if (mx%2==1) {
				For(i,mx-1) 
				{
					if (i&1) {For(j,m-1) putchar('R');}
					else {For(j,m-1) putchar('L'); }
					if (i<n) putchar('D');
				}

				int tx=mx,ty=1;
				int p=0;
				For(j,m)
				{
					if (my==j) {if (j<m) putchar('R');continue;}
					if (p==0) printf("D");
					else printf("U");
					p^=1;
					if (j<m) putchar('R');
				}
				Fork(i,mx+2,n) 
				{
					putchar('D');
					if ((i&1)^1) {For(j,m-1) putchar('R');}
					else {For(j,m-1) putchar('L'); }
				}
			}
			
			if (my%2==1) {
				For(i,my-1) 
				{
					if (i&1) {For(j,n-1) putchar('D'); }
					else {For(j,n-1) putchar('U'); }
					if (i<m) putchar('R');
				}
			
				int tx=1,ty=my;
				int p=0;
				For(j,n)
				{
					if (mx==j) {if (j<n) putchar('D');continue;}
					if (p==0) printf("R");
					else printf("L");
					p^=1;
					if (j<n) putchar('D');
				}
				
				Fork(i,my+2,m) 
				{
					putchar('R');
					if ((i&1)^1) {For(j,n-1) putchar('D'); }
					else {For(j,n-1) putchar('U'); }
				}
			}
			
			
		} 
		else {
			cout<<sum<<endl;
			if (n%2) {
				For(i,n) 
				{
					if (i&1) {For(j,m-1) putchar('R');}
					else {For(j,m-1) putchar('L'); }
					if (i<n) putchar('D');
				}
			} else {
				For(i,m) 
				{
					if (i&1) {For(j,n-1) putchar('D'); }
					else {For(j,n-1) putchar('U'); }
					if (i<m) putchar('R');
				}
			}
		}
		cout<<endl;
	}
	
	return 0;
}