8
18
2015
1

2015 Multi-University Training Contest 9 T7

啊。这道题好大。听说是杜教出的。。%%%%

题目如下:

Travelling Salesman Problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 34    Accepted Submission(s): 2
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

题目简述:从[1,1]到[n,m],每个格子只能进入一次,使进入的格子权值之和最大

这道题经过精密的模拟测试,发现当n或m为奇数时可以全都跑遍。
然而当n,m均为偶数时,a[i][j](i+j为奇数时的格子可以被单独绕过,i+j为偶数时他必定会牵连到其他格子,为了使答案最大,那么我们就要选取i+j为奇数的格子中权值最小的。

于是这道题就解决了。

 

#define _CRT_SECURE_NO_WARNINGS

#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>

using namespace std;

typedef long long LL;

void Main();
int main()
{
	Main();
	return 0;
}
int n, m;
int a[110][110];
int tot, minn;
int xx, yy;

void Main()
{
	while (scanf("%d%d", &n, &m) != EOF)
	{
		tot = 0; minn = 10010;
		for (int i = 1;i <= n;++i)
			for (int j = 1;j <= m;++j)
			{
				scanf("%d", &a[i][j]);
				tot += a[i][j];
				if ((i + j) & 1)
				{
					if (minn > a[i][j])
					{
						minn = a[i][j];
						xx = i; yy = j;
					}
				}
			}
		if (n & 1)
		{
			printf("%d\n", tot);
			for (int i = 1;i <= n / 2;++i)
			{
				for (int j = 1;j <= m - 1;++j)printf("R");
				printf("D");
				for (int j = 1;j <= m - 1;++j)printf("L");
				printf("D");
			}
			for (int j = 1;j <= m - 1;++j)printf("R");
			puts("");
			continue;
		}
		if (m & 1)
		{
			printf("%d\n", tot);
			for (int i = 1;i <= m / 2;++i)
			{
				for (int j = 1;j <= n - 1;++j)printf("D");
				printf("R");
				for (int j = 1;j <= n - 1;++j)printf("U");
				printf("R");
			}
			for (int j = 1;j <= n - 1;++j)printf("D");
			puts("");
			continue;
		}
		if (yy & 1)
		{
			printf("%d\n", tot - minn);
			for (int i = 1;i <= ((yy - 1) >> 1);++i)
			{
				for (int j = 1;j <= n - 1;++j)printf("D");
				printf("R");
				for (int j = 1;j <= n - 1;++j)printf("U");
				printf("R");
			}
			printf("R");
			for (int i = 1;i <= ((xx - 1) >> 1);++i)printf("DLDR");
			printf("D");
			for (int i = 1;i <= (n - xx) / 2;++i)printf("DLDR");
			for (int i = 1;i <= (m - yy) / 2;++i)
			{
				printf("R");
				for (int j = 1;j <= n - 1;++j)printf("U");
				printf("R");
				for (int j = 1;j <= n - 1;++j)printf("D");
			}
			puts("");
		}
		else
		{
			printf("%d\n", tot - minn);
			for (int i = 1;i <= ((yy - 1) >> 1);++i)
			{
				for (int j = 1;j <= n - 1;++j)printf("D");
				printf("R");
				for (int j = 1;j <= n - 1;++j)printf("U");
				printf("R");
			}
			for (int i = 1;i <= ((xx - 1) >> 1);++i)printf("RDLD");
			printf("D");
			for (int i = 1;i <= ((n - xx - 1) >> 1);++i)printf("RDLD");
			printf("R");
			for (int i = 1;i <= (m - yy) / 2;++i)
			{
				printf("R");
				for (int j = 1;j <= n - 1;++j)printf("U");
				printf("R");
				for (int j = 1;j <= n - 1;++j)printf("D");
			}
			puts("");
		}
	}
	return;
}
Category: 程序 | Tags: | Read Count: 260

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