2019 cattle off more school C Governing sand

The meaning of problems

There are $ n $ the number of trees, trees have a high degree per $ H_i $, costs $ C_i $, the number of $ P_i $, now you want to cut down some trees, so that the rest of the tree is more than the tallest tree in general, for the Minimum cost of. ($ 1 \ leq n \ leq 10 ^ 5, \ 1 \ leq H_i \ leq 10 ^ 9, \ 1 \ leq C_i \ leq 200, \ 1 \ leq P_i \ leq 10 ^ 9 $)

analysis

First of all, easy to think we should enumerate the height of the tallest trees, cut down all the higher, then select from the low cost of the k-th smallest cut.

It focuses on how to obtain the current minimum cost of k.

$ C_i $ notes that the range is very small, so we can count the cost of the number $ i $ the tree, denoted by $ C [i] $ (equivalent to the establishment of 200 barrels).

From low to high by highly sorted, traversing it again, get the global minimum cost. Complexity is $ O (200n) $, if the binary search can do $ O (log200 \ cdot n) $.

#include<bits/stdc++.h>
using namespace std;
#define LL long long
struct tree{
    LL h,c,p;
}a[100005];
bool cmp(tree a,tree b){
    return a.h<b.h;
}
LL n;
LL tmp=0,ans=0,tot=0;
LL num[205];
int main(){
    while(cin>>n){
        memset(num,0,sizeof(num));
        ans=tmp=tot=0;
        for(LL i=0;i<n;++i){
            scanf("%lld%lld%lld",&a[i].h,&a[i].c,&a[i].p);
            tmp+=a[i].c*a[i].p;
        }
        ans=tmp;
        sort(a,a+n,cmp);
        for(LL i=0,j;i<n;i=j){
            j=i;while(a[j].h==a[i].h&&j<n)++j;
            LL sum=0;
            LL cost=0;
            for(LL t=i;t<j;++t){
                tmp-=a[t].c*a[t].p;
                tot+=a[t].p;
                sum+=a[t].p;
            }
            sum=tot-sum*2+1;
            for(LL w=1;sum>0;++w){
                if(num[w]<=sum){
                    cost+=w*num[w];
                    sum-=num[w];
                }else{
                    cost+=sum*w;
                    break;
                }
            }
            ans=min(ans,cost+tmp);
            for(LL t=i;t<j;++t){
                num[a[t].c]+=a[t].p;
            }
        }
        cout<<ans<<endl;
    }
}

 

 

Reference Links: https://ac.nowcoder.com/acm/contest/view-submission?submissionId=41065020