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
乍一看，还以为是难题，解题思路很简单，每个点负责一条边，如果n,m中有一个偶数就用n*m就ok了，如果都是奇数，就要过一次斜边，用斜边长度根号二减去一加上n*m就行了。
#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
int main()
{
int cas;
cin>>cas;
int count=0;
while(cas--)
{
count++;
int n,m;
cin>>n>>m;
cout<<"Scenario #"<<count<<":"<<endl;
if(n%2==0||m%2==0)
printf("%.2f
",(double)n*m);
else
printf("%.2f
",(double)n*m-1+sqrt(2.0));
}
}