ZOJ P1037 HDOJ P1046 Gridland

Gridland

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2384    Accepted Submission(s): 1139

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

 

Source

Northwestern Europe 2001

网上找不到翻译  囧。。。。。

数学题，要找规律。

首先看当m或n中有一个为2的情况，显然，只需要算周长就ＯＫ了。即（ｍ+n-2）*2，考虑到至少其中一个为2，所以答案为2*m或2*n，亦即m*n。注意这里保证了其中一个数位偶数。
当ｍ，ｎ≥３时，考虑至少其中一个为偶数的情况，显然，这种情况很简单，可以得出，结果为m*n，又可以和上面这种情况合并。
 
下面看m，n均为奇数的情况，由于不好贴图，且这题又比较简单，就不多说了，就写个最后的公式吧：
（m+n-2）*2-1+sqrt（2）+（n-3）*(m-2)+(m-2)=m*n-1+sqrt(2)
本来还要考虑m，n互换后求最小值的，看到最后的公式后，也就不用再算了。考虑到题目里面要求保留两位小数，所以最终，公式直接定为：m*n+0.14

话说  C++输入输出流真心麻烦。

C++语言: ZOJ 1037 HDOJ 1046 Gridland.cpp by--小KD

// ZOJ 1037  HDOJ 1046#include<iostream>using namespace std;int main(){int n,a,b,i;cin>>n;cout.setf(ios_base::showpoint);cout.setf(ios_base::fixed);cout.precision(2);for(i=1;i<=n;++i){cin>>a>>b;cout<<"Scenario #"<<i<<":"<<endl;if(a%2 && b%2)cout<<double(a*b+0.4142)<<endl;elsecout<<double(a*b)<<endl; cout<<endl;}return 0;}

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