Bzoj1584: [usaco 2009 Mar] cleaning up cleaning

 

Description

 

There are native cows, each of which has a PI, 1 <= pI <= m <= n <= 40000. Now Farmer John is going to divide these cows into several sections and define the non-crab degree of each section as: if there are k different numbers in this section, then the non-crab degree is K * K. Then the total "no crab" degree is the sum of "no crab degree" in all segments.

Question: http://www.lydsy.com/JudgeOnline/problem.php? Id = 1584

Question:

Comparison of God's DP... How have we met God's questions recently...

Write some comments in the code

1 # include <cstdio> 2 3 # include <cstdlib> 4 5 # include <cmath> 6 7 # include <cstring> 8 9 # include <algorithm> 10 11 # include <iostream> 12 13 # include <vector> 14 15 # include <map> 16 17 # include <set> 18 19 # include <queue> 20 21 # include <string> 22 23 # define INF 100000000024 25 # define maxn 5000026 27 # define maxm 500 + 10028 29 # define EPS 1e-1030 31 # define ll long long32 33 # define PA pair <int, int> 34 35 # define for0 (I, n) for (INT I = 0; I <= (n); I ++) 36 37 # define for1 (I, n) for (INT I = 1; I <= (n); I ++) 38 39 # define for2 (I, x, y) for (INT I = (X ); I <= (y); I ++) 40 41 # define for3 (I, x, y) for (INT I = (x); I> = (y ); I --) 42 43 # define mod 100000000744 45 using namespace STD; 46 47 inline int read () 48 49 {50 51 int x = 0, F = 1; char CH = getchar (); 52 53 While (CH <'0' | ch> '9') {If (CH = '-') F =-1; ch = getchar ();} 54 while (CH> = '0' & Ch <= '9') {x = 10 * x + CH-'0 '; ch = getchar ();} 56 57 Return x * F; 58 59} 60 int n, m, a [maxn], B [maxn], C [maxn], f [maxn], pre [maxn]; 61 62 int main () 63 64 {65 66 freopen ("input.txt", "r", stdin ); 67 68 freopen ("output.txt", "W", stdout); 69 70 N = read (); M = SQRT (N ); 71 for1 (I, n) A [I] = read (); 72 memset (F, 60, sizeof (f); 73 memset (PRE,-1, sizeof (pre); // pre [I] indicates the position where I appeared in the previous sequence 74 f [0] = 0; 75 for1 (I, n) 76 {77 for1 (J, m) if (pre [A [I] <= B [J]) C [J] ++; // B [J] indicates that B [J] + 1 to I has <= J different numbers 78 pre [A [I] = I; 79 for1 (J, m) 80 If (C [J]> J) // C [J] indicates the number of different numbers from B [J] + 1 to I, if J is clearly not defined in B [J], delete 81 {82 int T = B [J] + 1; // pointer 83 while (pre [A [T]> T) T ++; // scan forward, pre [A [T]> T indicates that there is still a [T] between T and I, so we continue to move forward, knowing that there is no successor to a number, meaning 84 B [J] = t after I; C [J] --; 85} 86 for1 (J, m) f [I] = min (F [I], f [B [J] + J * j); 87} 88 printf ("% d \ n", F [N]); 89 90 return 0; 91 92}

View code

 

 

 

 

 

 

 

 

Bzoj1584: [usaco 2009 Mar] cleaning up cleaning