Hdu-3916-Sequence Decomposition

Question: a sequence is actually a function. The subscript starts from 1 and splits this function into the sum of several switching functions G (x, y). There can be multiple methods, calculate the maximum value of y-x + 1 in each method. --> This question is greedy! For: 2 2 2 obviously, the minimum length is y-x + 1 = 3 when it is divided into 3G (1, 3). If not, then, the minimum length must be smaller than 3 -- get -- when the two numbers are equal, they should be put into a G. For: 2 4 2 If Step 2, 4 is returned to a G instead of 2, 2G (1, 2) + 2G (2, 3), the minimum length is 2 at this time. If the three numbers at the beginning are classified into one G, 2G (1, 3) + 2G (2, 2) is obtained ), at this time, the minimum length is 1 -- available -- the minimum length is increased to 1G, and the minimum length is reduced to 1G by hours. At the beginning, I saved the input data to array a, for a's a [I] Again and again --, until all a [I] = 0, the result does not need to be said -- TLE; then, we can see that for 2 3 3 7 4 1, a [1] to a [4] can be directly-2 (the first smallest number on the left 2), for 4 4 4 7 4 1, a [1] to a [4] can be directly-3 (the largest number on the right-the number on the right). In this way, all a [I] can be reduced to 0, the result is the same-TLE; you can think of it. You cannot directly reduce all a [I] to 0 !!! --> Finally, consider G (L, R) and use m [R] to represent the amount to be subtracted from a [L] to a [R]. Scan, maintain m and the temp to be subtracted from the minimum number on the left, which is similar to the downloading mechanism of the line tree. [Cpp] # include <cstdio> # include <algorithm> # include <cstring> using namespace std; const int maxn = 10000 + 10; int main () {int T, N, i; long a [maxn], m [maxn]; // a is the input array, m [R] indicates the amount between a [L] And a [R] scanf ("% d", & T); while (T --) {scanf ("% d", & N); for (I = 1; I <= N; I ++) scanf ("% I64d ", & a [I]); a [N + 1] = 0; // Add 0 at the end. Handle int L = 1, R = 1, temp = 0, minn = 2147483647; // G (L, R), temp is the amount to be subtracted according to the specified a [L]. minn is the result. Memset (m, 0, sizeof (m); while (L <= N) {for (; L <= N; L ++) if (! (A [L]-temp) temp-= m [L]; // update else break; // find the first a [L] if (L> N) continue that is not 0; // it is possible to complete if (R <L) R = L, for example, 2 2 2 3 3 for (; R <= N; R ++) if (a [R + 1] <a [R]-m [R]) // find the right boundary {www.2cto.com minn = min (minn, R-L + 1 ); // update long dis = a [R]-m [R]-a [R + 1]; // dis G (L, R) dis = min (dis, a [L]-temp); m [R] + = dis; // accumulate incremental temp + = dis; // update break ;}} printf ("% d \ n", minn);} return 0 ;}