Triangular Pastures

Total Submission(s) : 94   Accepted Submission(s) : 33

Problem 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

511334

 

 

Sample Output

692

 

 

package DynamicProgrammingPackage;import java.util.Arrays;import java.util.Scanner;/* * 二维0-1背包问题 * 问题描述：给你一些边求这些栅栏，然后用这些栅栏构造一个三角形（栅栏不能截断）， *        求能组成的最大三角形的面积 * 思路： *    状态：v[i][j]==true代表根据这些栅栏能够构造出长度为i和j的栅栏 *    转移方程：v[i][j]=v[i-h[x]][j] || v[i][j-h[x]] *           因为如果能v[i-h[x]][j]或者v[i][j-h[x]]如果为true, *           则加上长度为h[x]的栅栏之后，就能够造出v[i][j] *    如何变成：用倒序的方式压缩空间 */public class TriangularPastures{//每个最大长度（40）*最大片段数（40）=三角形的最大面积（1600），所以三角形的最大边长是800public static final int N = 801;//背包的第一维和第二维分别是三角形的两条边public static boolean[][] v = new boolean[N][N];//背包容量public static int m;//一共有多少个栅栏public static int L;//三角形的周长public static int half;//三角形周长的一半public static int h[] = new int[40];public static void main(String[] args) {Scanner sc = new Scanner(System.in);while(sc.hasNextInt()) {m = sc.nextInt();L = 0;for(int i=0; i<m; i++) {h[i] = sc.nextInt();L += h[i];}half = L >> 1;fillBack(half+1);dp();double result = maxArea();if(result < 0) {System.out.println("-1");} else {System.out.printf("%d\n", (int)(Math.sqrt(result)*100));}}}private static double maxArea() {double maxArea = -1;double tempArea = 0;//开始的时候用的是L >> 1,一直Wrong answerdouble half1 = L / 2.0;int a = 0;for(int i=1; i<half; i++) {for(int j=1; j<half; j++) {if(v[i][j] == true) {a = L - i - j;if(i+j>a && i+a>j && j+a>i) {//判断两边之和大于第三边//三角形的面积=√L/2*(L/2-a)*(L/2-b)*(L/2-c),这里没有开根号，等输出的时候再开根号tempArea = half1 * (half1-a) * (half1-i) * (half1-j);maxArea = Math.max(tempArea, maxArea);}}}}return maxArea;}/** * 二维0-1背包 */private static void dp() {v[0][0] = true;for(int i=0; i<m; i++) {for(int j=half; j>=0; j--) {//为了压缩空间，所以倒着求for(int k=j; k>=0; k--) {//不用判断（j>=h[i] && v[j-h[i]][k]==true）,因为v[j-h[i]][k]原本就是falseif(j>=h[i]) {v[j][k] = v[j-h[i]][k];}if(k>=h[i]) {v[j][k] = v[j][k-h[i]];}}}}}/** * 初始化数组 * @param max 输入的栅栏数目 */private static void fillBack(int max) {for(int i=0; i<max; i++) {Arrays.fill(v[i], false);}}}

	
				
	
					   
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