hdu5135 Little Zu Chongzhi's Triangles

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1165    Accepted Submission(s): 655

Problem Description

Zu Chongzhi (429–500) was a prominent Chinese mathematician and astronomer during the Liu Song and Southern Qi Dynasties. Zu calculated the value ofπ to the precision of six decimal places and for a thousand years thereafter no subsequent mathematician computed a value this precise. Zu calculated one year as 365.24281481 days, which is very close to 365.24219878 days as we know today. He also worked on deducing the formula for the volume of a sphere. 

It is said in some legend story books that when Zu was a little boy, he liked mathematical games. One day, his father gave him some wood sticks as toys. Zu Chongzhi found a interesting problem using them. He wanted to make some triangles by those sticks, and he wanted the total area of all triangles he made to be as large as possible. The rules were :

1) A triangle could only consist of 3 sticks.
2) A triangle's vertexes must be end points of sticks. A triangle's vertex couldn't be in the middle of a stick.
3) Zu didn't have to use all sticks.

Unfortunately, Zu didn't solve that problem because it was an algorithm problem rather than a mathematical problem. You can't solve that problem without a computer if there are too many sticks. So please bring your computer and go back to Zu's time to help him so that maybe you can change the history.

 

Input

There are no more than 10 test cases. For each case:

The first line is an integer N(3 <= N<= 12), indicating the number of sticks Zu Chongzhi had got. The second line contains N integers, meaning the length of N sticks. The length of a stick is no more than 100. The input ends with N = 0.

 

Output

For each test case, output the maximum total area of triangles Zu could make. Round the result to 2 digits after decimal point. If Zu couldn't make any triangle, print 0.00 .

 

Sample Input

   
   
   
   

    
    
    
    3
1 1 20
7
3 4 5 3 4 5 90
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0.00
13.64
   
   
   
   

 

这题可以用状压dp做，用dp[state]表示该状态下组成的木棒的最大面积。那么我们对木棒数大于等于3的状态，可以加一个没有用过的木棒，然后枚举已经存在的两根木棒，把它们提出来和这根木棒结合，那么状态转移方程就是dp[state2]=max(dp[state2],dp[state1  ]+mianji(a[i],a[j],a[k]  )  );

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
double a[20];
double dp[10000];
int wei[20],tot,yigeshu;
void zhuanhuan(int state)
{
    int i,j;
    tot=0;
    yigeshu=0;
    while(state){
        wei[++tot]=state%2;
        if(state&1)yigeshu++;
        state>>=1;
    }
}

double mianji(double a,double b,double c)
{
    double d[5];
    d[1]=a;d[2]=b;d[3]=c;
    sort(d+1,d+4);
    if(d[1]+d[2]<=d[3])return 0;
    double p=(a+b+c)/2;
    return sqrt(p*(p-a)*(p-b)*(p-c));
}

double jisuan(int wei[])
{
    int i,j;
    int num=0;
    double c[10];
    for(i=1;i<=tot;i++){
        if(wei[i]){
            c[++num]=a[i];
        }
    }
    sort(c+1,c+4);
    return mianji(c[1],c[2],c[3]);
}


int main()
{
    int n,m,i,j,state,state1,state2,k;
    while(scanf("%d",&n)!=EOF && n!=0)
    {
        for(i=1;i<=n;i++){
            scanf("%lf",&a[i]);
        }

        memset(dp,0,sizeof(dp));
        for(state=1;state<=( (1<<n)-1 );state++ ){
            zhuanhuan(state);
            if(yigeshu==1 || yigeshu==2){
                dp[state]=0;continue;
            }
            if(yigeshu==3){
                dp[state]=jisuan(wei);
            }
            for(i=1;i<=n;i++){
                if( (state&(1<<(i-1)) )==0  ){
                    state2=( state|(1<<(i-1)) );
                    for(j=1;j<tot;j++){
                        for(k=j+1;k<=tot;k++){
                            if(wei[j] && wei[k]){
                                state1=(state^(1<<(j-1) ) );
                                state1=(state1^(1<<(k-1) ) );
                                dp[state2]=max(dp[state2],dp[state1  ]+mianji(a[i],a[j],a[k]  )  );
                            }
                        }
                    }
                }
            }
        }
        double maxnum=0;
        for(state=1;state<=( (1<<n)-1 );state++ ){
            maxnum=max(maxnum,dp[state]);
        }
        printf("%.2f\n",maxnum);
    }
    return 0;
}