Sgu 183. Painting the bils

Time Limit: 0.25 s

Space limit: 4 m

Question:

In N (n <= 10000) balls, color several balls at the cost of Ci, making any continuous M (M <= 100) at least two balls are colored.

 

 

 

 

 

 

 

Solution:

First, we can think of a DP state transition equation directly.

F [I] [J] indicates that the first I ball is applied and the last colored ball is the minimum cost of J.

F [I] [J] = min (F [J] [k]) + CI, k> I + 1-m

The time complexity of analyzing this transition equation is O (n * m). The data range of this question is as high as 10 ^ 8.

Obviously, we need a better solution.

By analyzing the above equation, we found that when calculating min (F [J] [k]), there is a part of repeated calculation,

So I tried to reduce this repetitive process.

For a J, the I range is (J + 1, J + s-1)

The range of corresponding K is (I + 1-m + 1 )~ J-1)

If we push I from (J + m-1) to (j + 1)

You can make K from (J-1) to (I + 1-m + 1)

Min (F [J] [k]) the range to be calculated increases sequentially and can be obtained recursively.

That is, min (F [J] [k]) can be obtained in O (1) time.

The total time complexity becomes O (N * m)

We can find that N * m arrays cannot be used directly in the space, and we can optimize the scrolling array.

Code

# Include <iostream> # include <cstring> using namespace STD; const int mod = 101; int n, m; int C [10009]; // F [I] [J] applies the current I-th ball, and the I-th, J-th ball at the minimum cost. // only the last 200 balls are retained; int f [200] [200]; int main () {CIN> N> m; For (INT I = 1; I <= N; I ++) cin> C [I]; memset (F, 0x3f, sizeof F); F [1] [0] = C [1]; for (INT I = 1; I <= m; I ++) for (Int J = 1; j <I; j ++) f [I] [J] = C [I] + C [J]; for (Int J = 2; j <n; j ++) {int TEM = 0x3f3f3f; for (INT I = J + m-1; I> J; I --) {if (I <= m) break; TEM = min (TEM, f [J % mod] [(I-m) % mod]); F [I % mod] [J % mod] = TEM + C [I];} int ans = 0x7fffffff; For (INT I = N-m + 1; I <= N; I ++) for (Int J = I-1; i-j <M & N-j <m; j --) ans = min (ANS, F [I % mod] [J % mod]); cout <ans; return 0 ;}

　　

Sgu 183. Painting the bils