HDU4666 + POJ2926 [far from manhattan]

At the beginning, I understood the N * 1 k algorithm,

The delete operation is still too time-consuming. Finally, I learned the corresponding STL set and map usage, which is much more convenient.

STL is really a good tool

 

#include<iostream>   #include<cstdio>   #include<map>   #include<set>   #include<vector>   #include<cstring>   using namespace std;  multiset<int> a[60005];  int x[60005][6];  int main()  {    int n,k,op,num;    while(scanf("%d%d",&n,&k)!=EOF)    {        for(int i=0;i<1<<k;i++)          a[i].clear();        for(int i=1;i<=n;i++)        {            scanf("%d",&op);            if(op==0)            {                for(int j=0;j<k;j++)                  scanf("%d",&x[i][j]);                for(int j=0;j<1<<k;j++)                {                   int s=0;                   for(int q=0;q<k;q++)                   {                       if(j&1<<q) s+=x[i][q];                       else s-=x[i][q];                   }                   a[j].insert(s);                }            }            else            {                scanf("%d",&num);                for(int j=0;j<1<<k;j++)                {                    int s=0;                    for(int q=0;q<k;q++)                    {                        if(j&1<<q) s+=x[num][q];                        else s-=x[num][q];                    }                    multiset<int>::iterator sum=a[j].find(s);                    a[j].erase(sum);                }            }            int ans=-100000000;            for(int j=0;j<1<<k;j++)            {                multiset<int>::iterator t=a[j].end();                t--;                int t1=(*t);                t=a[j].begin();                int t2=(*t);                ans=max(ans,t1-t2);            }            printf("%d\n",ans);        }    }    return 0;  }