2019牛客（第七场）C.Governing sand

题目链接、

题意：

给你n种树，每种树有p[i]个，每个去掉花费c[i]，树高h[i]

让你去掉一些树，使得最高的树的个数占总数的一半以上

题解：

从高到底枚举树（相同高度看为一种），计算出需要去掉多少，然后二分即可

 

一直错的原因：top=Q.top()忘记写了、

 

#include
using namespace std;
typedef long long ll;
const int maxn = 1e5+7;
struct node{
    ll P,C,H;//count,cost,height
    bool operator < (const node f)const{
        return C >Q;
        ll last = 0,Ans = 2e18;
        for(ll i=1;i<=n;i++){
            add(i,a[i].P,a[i].P*a[i].C);
            Q.push(make_pair(a[i].H,i));
        }
        while(!Q.empty()){
            ll Sum = 0,Cst = 0;
            pair top = Q.top();
            Q.pop();
            ll pos = top.second;
            Sum += a[pos].P;
            Cst += a[pos].C*a[pos].P;
            add(pos,-a[pos].P,-a[pos].P*a[pos].C);
            while(!Q.empty() && top.first == Q.top().first){
                top = Q.top();
                pos = top.second;
                add(pos,-a[pos].P,-a[pos].P*a[pos].C);
                Sum += a[pos].P;
                Cst += a[pos].C*a[pos].P;
                Q.pop();
            }
            ll Need = query(n)-Sum+1,cost = last;
            //printf("%lld %lld %lld\n",query(n),Sum,Need);
            if(Need>0){
                ll l = 1,r = n,sul = n;
                while(l<=r){
                    ll mid = l+r>>1;
                    if(query(mid)>=Need)
                        r = mid-1,sul = mid;
                    else
                        l = mid+1;
                }
                //printf("sul : %d \n",sul);
                ll cnt = Need-query(sul);
                cost += queryValue(sul-1)+(a[sul].P+cnt)*a[sul].C;
            }
            Ans = min(Ans,cost);
            last += Cst;
        }
        printf("%lld\n",Ans);
    }
    return 0;
}