Arthur and Table CodeForces, Arthur urcodeforces

Arthur and Table CodeForces, Arthur urcodeforces

Arthur and Table CodeForces-557C

First, sort by length.

The table legs with a length of p have a [p.

To make the table legs with a length of p the longest, cut down the sum {Number of legs with a length less than p}-a [p] + 1 leg at the cost. You also need to beam all the table legs longer than p. Enumerate p.

Key points (I can understand it after reading the question): the time/space complexity can be greatly reduced through a condition with a maximum of 200 energy.

Well, so... is it greedy?

 1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 struct Leg 5 { 6     int len,cost,len2; 7     bool operator<(const Leg& b) const 8     { 9         return len<b.len||(len==b.len&&cost<b.cost);10     }11 }l[100100];12 int l2[210];13 int sum1,n,ans=0x3f3f3f3f,now,now2,now3,sum2;14 int main()15 {16     int i,j;17     scanf("%d",&n);18     for(i=1;i<=n;i++)19         scanf("%d",&l[i].len);20     for(i=1;i<=n;i++)21         scanf("%d",&l[i].cost);22     sort(l+1,l+n+1);23     for(i=1;i<=n;i++)24         if(l[i].len!=l[i-1].len)25             l[i].len2=l[i-1].len2+1;26         else27             l[i].len2=l[i-1].len2;28     for(i=1;i<=n;i++)29         sum2+=l[i].cost;30     for(i=1;i<=n;i++)31     {32         sum2-=l[i].cost;33         now++;34         if(l[i].len2==l[i+1].len2)    continue;35         now2=sum1-now+1;36         if(now2<=0)37         {38             ans=min(sum2,ans);39         }40         else41         {42             now3=0;43             for(j=0;j<=200;j++)44             {45                 if(l2[j]>=now2)46                 {47                     now3+=j*now2;48                     break;49                 }50                 now2-=l2[j];51                 now3+=j*l2[j];52             }53             ans=min(now3+sum2,ans);54         }55         for(j=i;l[j].len2==l[j-1].len2;j--)56             l2[l[j].cost]++;57         l2[l[j].cost]++;58         sum1+=now;59         now=0;60     }61     printf("%d",ans);62     return 0;63 }