HDU5697 刷题计划 dp+最小乘积生成树

分析：就是不断递归寻找靠近边界的最优解

学习博客（必须先看这个）：

1：http://www.cnblogs.com/autsky-jadek/p/3959446.html

2：http://blog.csdn.net/u013849646/article/details/51524748

注：这里用的最小乘积生成树的思想，和dp结合

     每次找满足条件的最优的点，只不过BZOJ裸题的满足条件是形成一棵树

     这个题是大于m，生成树借用最小生成树进行求解最优，大于m用dp进行求解最优

#include <stdio.h>
#include <algorithm>
using namespace std;
const int N = 8e2+5;
typedef long long LL;
struct point{
  int x, y;
  point(int x=0,int y=0):x(x),y(y){}
  point operator -(const point &rhs)const{
     return point(x-rhs.x,y-rhs.y);
  }
  point operator +(const point &rhs)const{
    return point(x+rhs.x,y+rhs.y);
  }
};
LL cross(point a,point b){
  return 1ll*a.x*b.y-1ll*a.y*b.x;
}
int n,m,sum;
int a[N],b[N],c[N];
point p[N];
LL dp[N],f[N],ans;
point get(){
   p[0]=point(0,0),dp[0]=0;
   for(int i=1;i<=sum;++i)dp[i]=1LL<<60;
   for(int i=1;i<=n;++i){
    for(int j=sum;j>=a[i];--j){
      if(dp[j]>dp[j-a[i]]+f[i]){
         dp[j]=dp[j-a[i]]+f[i];
         p[j]=p[j-a[i]]+point(b[i],c[i]); 
      }
    }
   }
   int ret=m;
   LL tot=1LL*p[ret].x*p[ret].y;
   for(int i=m+1;i<=sum;++i){
     if(dp[i]<dp[ret]||dp[i]==dp[ret]&&1ll*p[i].x*p[i].y<tot)
      ret=i,tot=1ll*p[i].x*p[i].y;
   }
   ans=min(ans,tot);
   return p[ret];
}
void solve(point A,point B){
  for(int i=1;i<=n;++i)
    f[i]=1ll*c[i]*(B.x-A.x)+1ll*b[i]*(A.y-B.y);
  point C=get();
  if(cross(B-A,C-A)>=0)return;
  solve(A,C);solve(C,B);
}
int main(){
  while(~scanf("%d%d",&n,&m)){
    sum=0;
    for(int i=1;i<=n;++i){
      scanf("%d%d%d",&a[i],&b[i],&c[i]);
      sum+=a[i];
    }
    ans=1LL<<60;
    for(int i=1;i<=n;++i)f[i]=b[i];
    point A=get();
    for(int i=1;i<=n;++i)f[i]=c[i];
    point B=get();
    solve(A,B);
    printf("%I64d
",ans);
  }
  return 0;
}