2.7 sample sum tro 長城 2.3 owin trac cal

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

3 3
2 3 3
3 3 3
3 3 2

Sample Output

25
RRDLLDRR

Author xudyh
Source 2015 Multi-University Training Contest 9
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當n,m有一個奇數時，‘S形’可全取。

否則至少要少取一個，

假設少取(mx,my) ，當mx+my為偶數時，必定有一個與(mx,my)相鄰的不能取

否則必能全取剩下的。（‘S形’+特判2行‘長城形’)

#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;
}

HDU 5402(Travelling Salesman Problem-構造矩陣對角最長不相交路徑)