hdu 3548 Enumerate the Triangles （ 优 化 ）

题 目 ：

Problem Description

Little E is doing geometry works. After drawing a lot of points on a plane, he want to enumerate all the triangles which the vertexes are three of the points to find out the one with minimum perimeter. Your task is to implement his work.

 

Input

The input contains several test cases. The first line of input contains only one integer denoting the number of test cases.
The first line of each test cases contains a single integer N, denoting the number of points. (3 <= N <= 1000)
Next N lines, each line contains two integer X and Y, denoting the coordinates of a point. (0 <= X, Y <= 1000)

 

Output

For each test cases, output the minimum perimeter, if no triangles exist, output "No Solution".

 

Sample Input

2

3

0 0

1 1

2 2

4

0 0

0 2

2 1

1 1

 

Sample Output

 

Case 1: No Solution

Case 2: 4.650

 

参考 ：http://blog.csdn.net/dgq8211/article/details/7748105

题意：

平面上有n(n<=1000)点，问组成的三角形中，周长最小是多少。

优化：

周长c=L1+L2+L3，所以推得c > 2Li，假设Li的端点为点a、b，则又有Li>=| Xa-Xb |，故c > 2*| Xa-Xb |。

 

代码：

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>

using namespace std;

struct Length
{
    int x,y;
}L[1005];

bool com(Length a,Length b)
{
    return a.x<b.x;
}

int main( ) {

    int N,n,flag,xx=0;
    double Min,s1,s2,s3;
    cin >>N;
    while(N--){
        cin >>n;
        Min=99999; flag=0;
        for( int i=1;i<=n;i++)
           scanf("%d %d",&L[i].x,&L[i].y);
        sort(L+1,L+n+1,com);
        for( int i=1;i<=n-2;i++){
            for( int j=i+1;j<=n-1;j++){
                if( Min<=2*(L[j].x-L[i].x) ) break;
                s1=sqrt( ( (L[i].x-L[j].x)*(L[i].x-L[j].x)+(L[i].y-L[j].y)*(L[i].y-L[j].y) )*1.0);
                if( Min<=2*s1 ) continue;
                for( int k=j+1;k<=n;k++){
                    int t1=(L[i].x-L[j].x)*(L[j].y-L[k].y), t2=(L[j].x-L[k].x)*(L[i].y-L[j].y);
                    if( t1==t2 ) continue;
                    flag=1;
                    s1=sqrt( ( (L[i].x-L[j].x)*(L[i].x-L[j].x)+(L[i].y-L[j].y)*(L[i].y-L[j].y) )*1.0);
                    s2=sqrt( ( (L[k].x-L[j].x)*(L[k].x-L[j].x)+(L[k].y-L[j].y)*(L[k].y-L[j].y) )*1.0);
                    s3=sqrt( ( (L[k].x-L[i].x)*(L[k].x-L[i].x)+(L[k].y-L[i].y)*(L[k].y-L[i].y) )*1.0);
                    Min= min( Min, s1+s2+s3 );
                }
            }
        }
        cout <<"Case " <<++xx <<": ";
        if( flag ) printf("%.3lf
",Min);
        else printf("No Solution
");
    }
}