hdu 1046 Gridland

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1046

 

题目描述：

Gridland

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2631    Accepted Submission(s): 1249

Problem Description

For years, computer scientists have been trying to find efficient solutions to different computing problems. For some of them efficient algorithms are already available, these are the “easy” problems like sorting, evaluating a polynomial or finding the shortest path in a graph. For the “hard” ones only exponential-time algorithms are known. The traveling-salesman problem belongs to this latter group. Given a set of N towns and roads between these towns, the problem is to compute the shortest path allowing a salesman to visit each of the towns once and only once and return to the starting point.

The president of Gridland has hired you to design a program that calculates the length of the shortest traveling-salesman tour for the towns in the country. In Gridland, there is one town at each of the points of a rectangular grid. Roads run from every town in the directions North, Northwest, West, Southwest, South, Southeast, East, and Northeast, provided that there is a neighbouring town in that direction. The distance between neighbouring towns in directions North–South or East–West is 1 unit. The length of the roads is measured by the Euclidean distance. For example, Figure 7 shows 2 × 3-Gridland, i.e., a rectangular grid of dimensions 2 by 3. In 2 × 3-Gridland, the shortest tour has length 6.

 

 

Input

The first line contains the number of scenarios.

For each scenario, the grid dimensions m and n will be given as two integer numbers in a single line, separated by a single blank, satisfying 1 < m < 50 and 1 < n < 50.

 

 

Output

The output for each scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. In the next line, print the length of the shortest traveling-salesman tour rounded to two decimal digits. The output for every scenario ends with a blank line.

 

 

Sample Input

  

   2
2 2
2 3
  

 

 

Sample Output

  

   Scenario #1:
4.00

Scenario #2:
6.00
  

 

 

题意：给出一个矩阵点阵问应最短路经从一个点经过所有点以此回到该起点的长度。

 

题解：这题画图就很容易理解了，可以先画出3*4、3*3、4*4的点阵图案，然后试着在上面用最短路经走一走，可以发现这样一个规律，当矩形点阵的长宽都是奇数时，最短路经中必定有一条斜线；而只要有一边是长或宽是偶数就可以都通过直线来完成最短路经，所以只需要判断一下两边的奇偶情况就能求最短路径了。

 

代码：

/*
acm
hdu:Gridland

take a picture for 4*4  3*3 and 3*4  grid that you'll know the key to the problem
*/
#include<stdio.h>
#include<stdlib.h>
#include<math.h>

int t=0,m=0,n=0;

/*for test*/
int test()
{
	return(0);
}

/*main process*/
int MainProc()
{
	scanf("%d",&t);
	int cases=0;
	while(t--)
	{
		scanf("%d%d",&m,&n);
		cases++;
		printf("Scenario #%d:\n",cases);
		if(m%2!=0&&n%2!=0)
		{
			printf("%.2f\n\n",m*n-1+sqrt(2));
		}
		else
		{
			printf("%.2f\n\n",(float)(m*n));
		}
	}
	return(0);
}

int main()
{
	MainProc();
	return(0);
}