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.

輸入

* 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.

輸出

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.

樣例輸入

5
1
1
3
3
4

樣例輸出

692

提示

which is 100x the area of an equilateral triangle with side length 4

題目大意：

給定幾條邊，用上全部的邊組成三角形，輸出最大的三角形面積，若不存在，輸出-1.
大體思路：

枚舉所有可能出現(xiàn)的邊的情況，在判斷是否能成為三角形，然后進(jìn)行處理

#include <bits/stdc++.h> using namespace std; const int N=1605;///40*40 int dp[N][N]={0};///dp[i][j]代表能否構(gòu)成i,j兩邊 int a[50]; int area(int a,int b,int c) {double p=(a+b+c)/2.0;return (int)100*sqrt(p*(p-a)*(p-b)*(p-c)); } int main() {int n,sum=0;cin>>n;for(int i=0;i<n;i++)cin>>a[i],sum+=a[i];dp[0][0]=1;for(int i=0;i<n;i++)for(int j=sum;j>=0;j--)for(int k=j;k>=0;k--)if(j>=a[i]&&dp[j-a[i]][k]||k>=a[i]&&dp[j][k-a[i]])dp[j][k]=1;int maxs=-1;for(int i=sum;i>0;i--)for(int j=i;j>0;j--){if(dp[i][j]){int k=sum-i-j;if(i<j+k||i+j>k)///判斷三邊是否組成三角形maxs=max(maxs,area(i,j,k));}}if(maxs==-1) cout<<"-1"<<'\n';else cout<<maxs<<'\n';return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/zdragon1104/p/8990133.html

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