POJ 1948 Triangular Pastures(二维背包) - ACway的专栏 - 博客频道 - CSDN.NET


POJ 1948 Triangular Pastures(二维背包) - ACway的专栏 - 博客频道 - CSDN.NET
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
5  1  1  3  3  4  
Sample Output
692  
Hint
[which is 100x the area of an equilateral triangle with side length 4] 

   题意:
  求n条线段构成的三角形的最大面积*100
  题解
  由于三角形有三个边,并且所有的线段都要用到,于是可以用dp[i][j]表示边长为i,j,sum-i-j的三角形,由于是二维的,j,k边有一个可以取到0,另一个减去a后要大于等于0,最后用海伦公式算面积。



http://shangxun.iteye.com/blog/1937614
f[i][j][k] 表示用前i条边,能否组成长度为j和k的两条边
初始化f[0][0][0] = true;
f[i][j][k] = f[i-1][j-len[i]][k] || f[i-1][j][k-len[i]];
如果f[i][j][k]=true,那么就计算以j,k,sum-j-k为三条边能否组成三角形,如果可以就用海伦公式计算面积。

由于如果要开三维数组的话,要开f[40][1600][1600],明显是要MLE的,所以要降成二维的,
而时间复杂度40*1600*1600,如果没有优化直接上,是要TLE的。
所以要优化,根据优化的程度,运行时间可以再100MS~1000MS之间:
1. f[i][j] 和 f[j][i]是相同的两个三角形,所以递推时可以让 j>=i
2. 对于每条边,一定是小于周长的一半
3. 枚举到第i条边时, 前i条边不可能组成大于等于sum[i] (sum[i]=len[1]+...+len[i])的长度
  1. inline bool isTriangle(double e[3]){  
  2.     if(e[2] < e[1]){  
  3.         swap(e[2], e[1]);  
  4.         if(e[1] < e[0]) swap(e[0], e[1]);  
  5.     }  
  6.     return e[0]+e[1] > e[2];  
  7. }  
  8.   
  9. inline double getArea(double e[3]){  
  10.     if(!isTriangle(e)) return -1;  
  11.     double p = sum[n]/2.0;  
  12.     return sqrt(p*(p-e[0])*(p-e[1])*(p-e[2]));  
  13. }  
  14.   
  15. int main(){  
  16.     scanf("%d", &n);  
  17.     for(int i=1; i<=n; ++i){  
  18.         scanf("%d", &len[i]);  
  19.         sum[i] = sum[i-1]+len[i];  
  20.     }  
  21.   
  22.     memset(f, 0, sizeof(f));  
  23.     f[0][0] = true;  
  24.   
  25.     double ans = -1;  
  26.     double e[3];  
  27.     for(int i=1; i<=n; ++i){  
  28.         for(int v1=min(sum[i],sum[n]>>1); v1>=0; --v1){  
  29.             for(int v2=min(sum[i]-v1,sum[n]>>1); v2>=v1; --v2)if(v1+v2<sum[n]){  
  30.                 e[0] = v1; e[1] =v2; e[2] = sum[n]-v1-v2;           
  31.                 if(v1-len[i]>= 0 && f[v1-len[i]][v2]){  
  32.                     f[v1][v2] = true;  
  33.                 }  
  34.                 if(!f[v1][v2] && v2-len[i]>=0 && f[v1][v2-len[i]]){  
  35.                     f[v1][v2] = true;  
  36.                 }  
  37.                 if(f[v1][v2]){  
  38.                     ans = max(ans, getArea(e));               
  39.                 }  
  40.             }   
  41.         }  
  42.     }  
  43.     if(ans < 0) puts("-1");  
  44.     else printf("%d\n", (int)(100*ans));  
  45.   
  46.     return 0;  
  47. }  
http://www.cnblogs.com/wuyuewoniu/p/4304797.html
先求面积,用海伦公式,s=sqrt(p*(p-a)*(p-b)*(p-c)),其中a,b,c分别为三角形的三条边,p为三角形的半周长,同时由这个根式可以推出,三角形的任意一条边小于其半周长(根号里面大于0,如果等于0面积为0没有意义了)
所以考虑背包两条边,再用周长减去这两条边求出第三条边,再遍历一遍找出最大的三角形。
dp[i][j]表示三角形的第一条边为i,第二条边为j所构成的三角形是否存在,
如果存在,dp[i][j]=1,否则为0 当dp[i][j]=1的时候,dp[i+l[k]][j]和dp[i][j+l[k]]也为1
19     while(scanf("%d",&n)!=EOF)
20     {
21         int sum=0, ans=-1;    
22         for(i=1;i<=n;i++) {scanf("%d",&l[i]);sum+=l[i];}
23         memset(dp,0,sizeof(dp));
24         dp[0][0]=1;
25         c=sum;
26         sum=c/2-(c/2==0);
27         
28         for(k=1;k<=n;k++)
29             for(i=sum;i>=0;i--)
30                 for(j=i;j>=0;j--)
31                     if(dp[i][j]) dp[i][j+l[k]]=dp[i+l[k]][j]=1;
32     
33         for(i=sum;i>=1;i--)
34             for(j=i;j>=1;j--)
35                 if(dp[i][j]) ans=max(ans,area(i,j,c-i-j));
36         printf("%d\n",ans);
37     }
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