detail:http://blog.csdn.net/lvshubao1314/article/details/41624239
Triangular Pastures
Time Limit: 1000MS | Memory Limit: 30000K | |
Total 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.
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.
* 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]
Source
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; }