POJ1948：Triangular Pastures（二维01背包）

detail：http://blog.csdn.net/lvshubao1314/article/details/41624239

Triangular Pastures

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 7802 Accepted: 2590

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

Hint

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

Source

USACO 2002 February

题意：给定N条木棒，求用完这些木棒能拼成的三角形面积最大值。

思路：范围只有40，可以先枚举所有情况，dp[i][j]表示含有i和j两条边的三角形，类似01背包：dp[i][j] = (dp[i][j-a[k]] || dp[i-a[k]][j])?1:0。

# include <stdio.h># include <math.h># include <string.h># include <algorithm>using namespace std;int dp[801][801], a[41];int main(){    int n, sum, half;    while(~scanf("%d",&n))    {        sum = 0;        for(int i=0; i<n; ++i)        {            scanf("%d",&a[i]);            sum += a[i];        }        half = sum >> 1;//优化：三角形边长小于等于周长一半。        memset(dp, 0, sizeof(dp));        dp[0][0] = 1;        for(int i=0; i<n; ++i)            for(int j=half; j>=0; --j)                for(int k=j; k>=0; --k)//优化：k等于j即可，因为dp[2][3]与dp[3][2]这种情况不必重复。                    if(j>=a[i]&&dp[j-a[i]][k] || k>=a[i]&&dp[j][k-a[i]])                        dp[j][k] = 1;        int imax = -1;        for(int i=half; i>=1; --i)            for(int j=i; j>=1; --j)            {                if(dp[i][j])                {                    int k = sum - i - j;                    if(i+j>k && i+k>j && j+k>i)                    {                        double tmp = (i+j+k)*1.0/2;                        int area = int(sqrt(tmp*(tmp-i)*(tmp-j)*(tmp-k))*100);//海伦公式                        imax = max(imax, area);                    }                }            }        printf("%d\n",imax);    }    return 0;}