Codeforces Round #437 (Div. 2, based on MemSQL Start[c]UP 3.0 - Round 2) C. Ordering Pizza

Posted 2020-10-09 早知如此绊人心，何如当初莫相识。

题意：有2种pizzaA和B，有n个人，每块pizz分为S片，给出每个人需要吃s片，吃一片A获得x的幸福，吃一片B获得y的幸福，问在购买最少的Pizza的前提下获得的最大幸福值

思路：注意每个人可同时吃A和B，当x>=y时，我们选择A，否则选B，看这个时候买的pizza,和所有片数买的pizza,否则的话，我们分开，对于多余的A，我们买B，用x-y最小的去换，B也同理

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int  N=1e5+100;
 5 
 6 struct node{
 7     ll s,x;
 8     node(ll xx,ll yy){
 9         s=xx;x=yy;
10     }
11 };
12 ll n,S;
13 vector<node> a,b;
14 ll sum1=0,sum2=0,sum=0;
15 bool cmp(node p,node q){
16     return p.x<q.x;
17 }
18 
19 bool check(){
20     ll x1=sum1/S,x2=sum2/S,x3=(sum1+sum2)/S;
21     if(sum1%S) x1++;
22     if(sum2%S) x2++;
23     if((sum1+sum2)%S) x3++;
24     if(x1+x2==x3) return true ;
25     else return false;
26 }
27 
28 int main(){
29     cin>>n>>S;
30     ll x,y,z;
31     for(int i=1;i<=n;i++){
32         scanf("%lld%lld%lld",&x,&y,&z);
33         if(y>=z){
34             a.push_back({x,y-z});
35             sum1+=x*1LL;
36             sum+=(x*y)*1LL;
37         }
38         else {
39             b.push_back({x,z-y});
40             sum2+=x*1LL;
41             sum+=(1LL)*x*z;
42         }
43     }
44     if(check()){
45         cout<<sum<<endl;return 0;
46     }
47     sort(a.begin(),a.end(),cmp);
48     sort(b.begin(),b.end(),cmp);
49    // cout<<sum<<" "<<sum1<<" "<<sum2<<endl;
50     ll ans1=sum,ans2=sum;
51     ll xx=sum1%S;
52     for(int i=0;i<a.size();i++){
53         ll Min=min(xx,a[i].s);
54         ans1-=Min*a[i].x;
55         xx-=Min;
56         if(xx==0) break;
57     }
58     ll yy=sum2%S;
59     for(int i=0;i<b.size();i++){
60         ll Min=min(yy,b[i].s);
61         ans2-=Min*b[i].x;
62         yy-=Min;
63         if(yy==0) break;
64     }
65     cout<<max(ans1,ans2)<<endl;
66 }