P2085 最小函数值

链接：Miku

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显然在x>0的时候，函数都是随x增大而增大的

所以说嘛，优先队列就能搞出来

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#include<iostream>
#include<cstdio> 
#include<algorithm>
#include<queue>
using namespace std;
int n,m;
int a[10001],b[10001],c[10001];
struct need{
    int x;
    int i;
    int v;
    friend bool operator < (need a,need b){
        return a.v>b.v;
    }
} now,x;
priority_queue <need>q ;
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;++i){
        scanf("%d%d%d",&a[i],&b[i],&c[i]);
        now.x=1;
        now.i=i;
        now.v=a[i]+b[i]+c[i];
        q.push(now);
    }
    for(int i=1;i<=m;++i){
        x=q.top();
        q.pop();
        cout<<x.v<<" ";
        now.i=x.i;
        now.x=x.x+1;
        now.v=a[now.i]*now.x*now.x+now.x*b[now.i]+c[now.i];
        q.push(now);
    }
    return 0;
}