zoukankan      html  css  js  c++  java
  • 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(1n,m100,nm2).

    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,y1), "R" means you walk to cell (x,y+1), "U" means you walk to cell (x1,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
     

    Recommend
    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;
    }
    







  • 相关阅读:
    删除名字和参数
    更改NX TITLE为路径
    我自己写的创建刀具
    创建刀具,很有用的信息
    控件改名
    已知体的最大尺寸在一堆实体里面找这个体
    cam对象类型
    ORACLE导入导出工具的使用
    ORACLE表空间
    Statement与PreparedStatement的区别
  • 原文地址:https://www.cnblogs.com/claireyuancy/p/7345404.html
Copyright © 2011-2022 走看看