Gridland
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5137 Accepted Submission(s):
2342
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.
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.
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
Source
Recommend
一道找规律的题，找到规律之后很简单。
题意：给定n*m的格点，问从一个角开始，遍历所有的点一次然后回到起点的最短距离。
附上代码：
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: ",j); 17 printf("%.2lf ",s); 18 j++; 19 } 20 return 0; 21 }