ZOJ_1037_Gridland

Gridland

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Time Limit: 2 Seconds       Memory Limit: 65536 KB

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Background

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.

Problem

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.

 
Figure 7: A traveling-salesman tour in 2 * 3-Gridland.

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：看图基本能明白意思了，就是经过所有的结点要产生回路，所有结点只经过一次。

思路：直接画图举几个例子很容易找到规律，m,n相乘为偶数答案就是m*n，如果为奇数，最后则需要走斜线答案就是，m*(n-1)+1.41(就是根号2)

Sample Program Here

#include<iostream>
#include<iomanip>  //控制格式的头文件
using namespace std;
///////////偶数就是m*n,奇数答案就是m*n+0.41注意最后转化成double输出 还有最后也是有空行的
int main(){

  int i=1,m,n,num;
  cout<<fixed<<setprecision(2);
  cin>>num;
  for(i=1;i<=num;i++){
	  cin>>m>>n;
	  if( (m*n)%2==0 ){
	     cout<<"Scenario #"<<i<<":"<<endl;
		 cout<<(double)m*n<<endl;
	  }
	  else{
	     cout<<"Scenario #"<<i<<":"<<endl;
		 cout<<m*n+0.41<<endl;	  
	  }
	  cout<<endl;
  }

  return 0;
}