zoj3399(DP+单调队列优化)

zoj3399(DP+单调队列优化)

题目链接： http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4007
很好的一道题目，用f[i,j]表示前i个学生分为j个班时的最小unhappy值，状态转移方程为：f[i,j]=min{f[k,j-1],+(sum[i]-sum[k])*g[j]}.其中a<=i-k<=b,sum[i] = sigma {(x[s] -L)^2 | s≤ i}.
但现在的时间复杂度为O(k*n^2)，显然会TLE。将转移方程整理一下：f[i,j]=sum[i]*g[j]+min{f[k,j-1]-sum[k]*g[j]}. 显然可维护一个单调递增队列，每次转移之前将新的可选状态从队列尾插入，同时保证队列头的元素在可选范围之内。这样每次转移就可以只取队列头的元素，总的时间复杂度为O(n*k).

#include < iostream >
using   namespace  std;
const   long   long  inf = 100000000000000ll;
long   long  sum[ 10005 ],aver;
long   long  f[ 10005 ][ 205 ];
int  x[ 10005 ],g[ 205 ];
int  n,m,a,b;
typedef  struct
{
   long long unhap;
   int loc;
} node;
node q[ 10005 ];

int  main()
{
    int test=0;
    while(scanf("%d%d%d%d",&n,&m,&a,&b)!=EOF)
    {
         test++;
         aver=0;
         int i,j;
         for(i=1;i<=n;i++)
         {
              scanf("%d",&x[i]);
              aver+=x[i];
         }
         aver/=n;
         for(i=1;i<=m;i++)
         scanf("%d",&g[i]);
         sum[0]=0;
         for(i=1;i<=n;i++)
         sum[i]=sum[i-1]+(x[i]-aver)*(x[i]-aver);
         for(i=0;i<=n;i++)
         for(j=0;j<=m;j++)
         f[i][j]=inf;
         f[0][0]=0;
         long long ans_unhap=inf;
         int ans_k=m+1,ans_t=n+1;
         for(j=1;j<=m;j++)
         {
              int head=0,tail=0;
              for(i=a;i<=n;i++)
              {
                   node temp;
                   if(f[i-a][j-1]!=inf)
                   {
                        temp.unhap=f[i-a][j-1]-sum[i-a]*g[j];
                        temp.loc=i-a;
                        while(head<tail&&q[tail-1].unhap>=temp.unhap) tail--;  //用>=是为了保证最后一个班的学生在多解的情况下人数最少。 
                        q[tail].unhap=temp.unhap;
                        q[tail++].loc=temp.loc;
                   }
                   while(head<tail&&q[head].loc<i-b) head++;
                   if(head<tail) f[i][j]=q[head].unhap+sum[i]*g[j];
              }
              if(f[n][j]!=inf)  //求多解情况下的最小分班数。 
              {
                   if(ans_unhap>f[n][j])
                   {
                        ans_unhap=f[n][j];
                        ans_k=j;
                        ans_t=n-q[head].loc;
                   }
              }
         }
         if(test!=1) printf("\n");
         if(ans_unhap==inf) printf("No solution.\n");
         else printf("%lld %d %d",ans_unhap,ans_k,ans_t);
    }
    return 0;
}