LS 21 Packing（DP）

Packing

The cost of set {(a1,b1),(a2,b2),…,(an,bn)} is defined as

max{ai}×max{bi}

Partition set {(a1,b1),(a2,b2),…,(an,bn)} into several non-empty subsets to minimize the sum of the cost of subsets.

Input

The first line contains an integer n.

n lines follow. Each of them contains two integer ai,bi.

(1≤n≤1000,1≤ai,bi≤1000)

Ouptut

Single integer denotes the minimum value.

Sample input

3 
1 1
1 3
3 1

Sample output

6

Note

Partition {{(1,1),(1,3)},{(3,1)}} gives the cost of 1×3+3×1=6.

思路：用数组maxn[x][y]表示x到y中最大的b;a就直接排序就行了

          dp[x]表示排序后的前x组的最优值。dp[x]=min(dp[x],maxn[j][i]*a[i]){j<=i}a[i]是排完序以后的。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int mm=1010;
const int oo=1e9;
class node
{
  public:int a,b;
}f[mm];
int dp[mm],maxn[mm][mm];
int n;
bool cmp(node a,node b)
{ if(a.a^b.a)
  return a.a<b.a;
  return a.b<b.b;
}
int main()
{
  while(cin>>n)
  {
    for(int i=1;i<=n;i++)
      cin>>f[i].a>>f[i].b;
    sort(f+1,f+n+1,cmp);///以a排序
    memset(maxn,0,sizeof(maxn));
    for(int i=1;i<=n;i++)///找最大的b
    { maxn[i][i]=f[i].b;
      for(int j=i+1;j<=n;j++)
      maxn[i][j]=maxn[i][j-1]>f[j].b?maxn[i][j-1]:f[j].b;
    }
    dp[0]=0;
    for(int i=1;i<=n;i++)
    { dp[i]=oo;int z;
      for(int j=1;j<=i;j++)
      { z=maxn[i-j+1][i]*f[i].a+dp[i-j];
        if(dp[i]>z)dp[i]=z;
      }
    }
    cout<<dp[n]<<"\n";
  }
}