POJ1948 Triangular Pastures(DP)

题目来源：
http://poj.org/problem?id=1948
Description
Like everyone, cows enjoy variety. Their current fancy is new shapes for pastures. The old rectangular shapes are out of favor; new geometries are the favorite. 
I. M. Hei, the lead cow pasture architect, is in charge of creating a triangular pasture surrounded by nice white fence rails. She is supplied with N (3 <= N <= 40) fence segments (each of integer length Li (1 <= Li <= 40) and must arrange them into a triangular pasture with the largest grazing area. Ms. Hei must use all the rails to create three sides of non-zero length. 
Help Ms. Hei convince the rest of the herd that plenty of grazing land will be available.Calculate the largest area that may be enclosed with a supplied set of fence segments. 
Input
* Line 1: A single integer N 
* Lines 2..N+1: N lines, each with a single integer representing one fence segment's length. The lengths are not necessarily unique. 
Output
A single line with the integer that is the truncated integer representation of the largest possible enclosed area multiplied by 100. Output -1 if no triangle of positive area may be constructed. 
Sample Input
5
1
1
3
3
4
Sample Output
692
Hint
[which is 100x the area of an equilateral triangle with side length 4] 
思路：
http://www.cnblogs.com/gj-Acit/p/3440507.html
题目大意就是说给你n个木棍让你用它们摆成一个三角形  使得这个三角形的面积最大。
先求出通过这n个木棍可以构成哪些边长组合的三角形，最后再一一比较这些三角形的面积。
代码：

（自己修改上面网址代码后的代码，便于理解）

#include <iostream>using namespace std;#include <cmath>bool DP[880][880]; //能够构成的两条边长 int len[42], tot, N;bool Judge(int a,int b,int c){if(a+b>c && a+c>b && b+c>a) return true;return false;}double area(int a, int b, int c){double p = (double)(a+b+c)/2.0;return sqrt(p*(p-a)*(p-b)*(p-c)); //求面积公式}int main(){int i,j,k;while(cin>>N){for(i=0;i<N;i++){cin>>len[i];tot += len[i];}int half = tot/2;DP[0][0] = 1;for( k=0;k<N;k++){for( i=half;i>=0;i--){ //边长小于总长一半,必须是从大到小进行，不然一条边会被放多次for( j=half;j>=0;j--){if((i-len[k]>=0 && DP[i-len[k]][j]) || (j-len[k]>=0 &&DP[i][j-len[k]])){DP[i][j]=1;}}}}double ans = -1;for( i=0;i<=half;i++)for( j=0;j<=half;j++)if(DP[i][j] && Judge(i,j,tot-i-j))ans = max(ans, 100*area(i,j,tot-i-j));cout<<(int)ans<<endl;}return 0;}