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.
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.
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.
We have carefully selected several similar problems for
you:
1038 1033 1036 1124 1129
1 #include <iostream>
2 #include <cstdio>
3 using namespace std;
4 int main()
5 {
6 int m,k,n,j=1;
7 double s;
8 scanf("%d",&k);
9 while(k--)
10 {
11 scanf("%d%d",&m,&n);
12 if(m%2==0||n%2==0)
13 s=m*n; //若n是偶数，我们可以将m的最后一列留出来和n的最下面一行，为了回来，而剩余的段走S形完全遍历；m是偶数时同理，共n*m。
14 else
15 s=m*n+0.414; //若都是奇数时，必须走一次斜线，才能变成上面的情况。共n*m-1+sqrt(2)。
16 printf("Scenario #%d:\n",j);
17 printf("%.2lf\n\n",s);
18 j++;
19 }
20 return 0;
21 }
