hdu-1046Gridland

图论规律（关于哈密尔顿图）

Gridland

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3283    Accepted Submission(s): 1520

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

22 22 3

 

Sample Output

Scenario #1:4.00Scenario #2:6.00

 

题目大意：就是给你一些点，而这些点是有规律的排列（如图），现在从一个点出发，遍历所有的点，最后回到起点，但是途中不能重复经过某些点，其实就是哈密尔顿回路。求最短路径。

解题思路：关于哈密尔顿回路有一个小规律，那就是当n.m至少有一个是偶数时，最短路径就是n*m。否则就是n*m-1+sqrt(2)。

#include<cstring>#include<cstdlib>#include<cstdio>#include<cmath>#include<iostream>#include<algorithm>using namespace std;int main(){    int t,n,m,i;    scanf("%d",&t);    for(i=1;i<=t;i++)    {        scanf("%d%d",&n,&m);        if((n%2==0)||(m%2==0))            printf("Scenario #%d:\n%d.00\n\n",i,n*m);        else            printf("Scenario #%d:\n%.2lf\n\n",i,n*m-1+sqrt(2.0));    }    return 0;}