[Luo Gu P1169] [explanations] [ZJOI2007] lunch

This is the subject of it?

Obviously the DP, speaking several important places

1. greedy: Let's eat a long time to line up (proof omitted)

2. Status:

f [i] [j] [k] i personal representative of the former, One time j, k is clearly time MLE II

So compressed into the front f [i] [j] i represents individuals One time j, with prefixes and maintenance time, s [i] -j II is the time

Code：

. 1 #include <bits / STDC ++ H.>
 2  the using  namespace STD;
 . 3  struct the Person {
 . 4      int Take, EAT;
 . 5      BOOL  operator <(the Person Ano) const {
 . 6          return EAT> ano.eat;
 . 7      }
 . 8 } P [ 233 ];
 . 9  int n-, ANS, S [ 233 ], F [ 233 ] [ 233 * 233 ]; // time to open square 
10 inline void the Init () {
 . 11      CIN >>n;
12     for(int i=1;i<=n;i++)cin>>p[i].take>>p[i].eat;
13     sort(p+1,p+1+n);//贪心 
14     //前缀和 
15     for(int i=1;i<=n;i++)s[i]=s[i-1]+p[i].take;
16     memset(f,0x3f,sizeof(f));
17     ans=0x3f3f3f3f;
18     f[0][0]=0;
19 }
20 inline void DP(){
21     for(int i=1;i<=n;i++){
22         for(int j=0;j<=s[i];j++){
23             //去一号窗口 
24             if(j>=p[i].take)f[i][j]=min(f[i][j],max(f[i-1][j-p[i].take],j+p[i].eat));
25             //去二号窗口 
26             f[i][j]=min(f[i][j],max(f[i-1][j],s[i]-j+p[i].eat));
27         }
28     }
29     for(int I = 0 ; I <= S [n-]; I ++) ANS = min (ANS, F [n-] [I]);
 30      COUT ANS << << endl;
 31 is  }
 32  int main () { // Ultra simple main 
33 is      the Init ();
 34 is      the DP ();
 35      return  0 ;
 36 }