Triangular Pastures

最新推荐文章于 2025-08-08 22:01:10 发布

转载最新推荐文章于 2025-08-08 22:01:10 发布 · 55 阅读

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原文链接：http://www.cnblogs.com/lavender913/p/3310813.html

文章标签：

#java

本文介绍了一个编程问题，任务是寻找由给定长度的围栏组成的最大三角形牧场面积。通过动态规划方法，文章提供了一段C++代码实现，用于解决此问题。

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Triangular Pastures

时间限制(普通/Java):3000MS/10000MS 运行内存限制:65536KByte

描述

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.

输入

* 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.

输出

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.

样例输入

样例输出

692

题目大意：给出n条边，找出可以拼成的一个三角形的最大面积

#include<iostream>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;
bool flag[810][810];

int main() {

    int n,i,j,k,ans;
    int str[110];

    while(~scanf("%d",&n)) {
        ans = 0;
        for(i=1; i<=n; i++) {
            scanf("%d",&str[i]);
            ans+=str[i];
        }
        memset(flag,false,sizeof(flag));
        flag[0][0]=true;
        int mid=ans/2;

　　　　　//判断是否可以形成三角形
        for(i=1; i<=n; i++)
            for(j=mid; j>=0; j--)
                for(k=j; k>=0; k--)
                    if(j>=str[i]&&flag[j-str[i]][k] || k>=str[i]&&flag[j][k-str[i]])
                        flag[j][k]=true;

        int Max=-1;
        for(i=mid; i>=1; i--)
            for(j=i; j>=1; j--) {
                if(flag[i][j]) {
                    k=ans-i-j;
                    if(i+j>k && i+k>j && j+k>i) {
                        double p=(i+j+k)/2.0;
                        int c=(int)(sqrt(p*(p-i)*(p-j)*(p-k))*100);
                        if(c>Max)
                            Max=c;
                    }
                }
            }
        printf("%d\n",Max);
    }
    return 0;
}