Sicily 1800 sequence (rmq)

// Rmq (solving a line segment tree) <br/> // after a, I understand the conversion of the series. If continuous intervals and, it must be converted to sum [I], indicating the sum of the intervals from 1 to I <br/> // This requires the sum of any intervals. Suppose [I, j] of the sum, only need sum [J]-sum [I-1] can be obtained immediately <br/> // note sum [0] = 0; <br/> // train of thought: Enumerate sum [I] (I> = L) from L) <br/> // find the maximum sum [J] (max (0, I-u) <= j <= I-l) in the range that meets the length condition) <br/> // sum [I]-sum [J] is the smallest <br/> // This converts the problem to the maximum value of the given range. <br/>/ /this question cannot be done using the st algorithm, if St is used, it will time out. It may take too much time in the log operation. <br/> // After Reading others' question solutions, we can see that the queue is optimized with priority, this speed should be faster <br/> # include <iostream> <br/> # define Max 32780 <br/> # define INF 2147483647 <br/> using namespace STD; <br/> int n, l, U; <br/> int sum [Max]; <br/> int _ max, ans; <br/> struct seg <br/> {<br/> int L, R, _ max; <br/>} segtree [Max * 4]; <br/> void buildtree (int fa, int L, int R) <br/>{< br/> segtree [fa]. L = L; <br/> segtree [fa]. R = r; <br/> If (L = r) <br/> {<br/> segtree [fa]. _ max = sum [l]; <br/> return; <br/>}< br/> int mid = (L + r)> 1; <br/> buildtree (2 * Fa, L, mid); <br/> buildtree (2 * Fa + 1, Mid + 1, R ); <br/> segtree [fa]. _ max = max (segtree [2 * Fa]. _ max, segtree [2 * Fa + 1]. _ max); <br/>}< br/> void search (int fa, int L, int R) <br/> {<br/> If (segtree [fa]. L = L & segtree [fa]. R = r) <br/>{< br/> _ max = max (_ max, segtree [fa]. _ max); <br/> return; <br/>}< br/> int mid = (segtree [fa]. L + segtree [fa]. r)> 1; <br/> If (r <= mid) Search (2 * Fa, L, R); <br/> else if (L> mid) search (2 * Fa + 1, L, R); <br/> else <br/>{< br/> Search (2 * Fa, L, mid ); <br/> Search (2 * Fa + 1, Mid + 1, R); <br/>}< br/> int main () <br/>{< br/> // freopen ("in.txt", "r", stdin); <br/> while (scanf ("% d", & N) & N) <br/>{< br/> ans = inf; <br/> scanf ("% d", & L, & U ); <br/> sum [0] = 0; <br/> for (INT I = 1; I <= N; ++ I) <br/>{< br/> scanf ("% d", Σ [I]); <br/> sum [I] + = sum [I-1]; <br/>}< br/> buildtree (1,0, n); <br/> for (INT I = L; I <= N; ++ I) <br/>{< br/> _ max = (-1) * INF; <br/> Search (1, max (0, I-U ), i-l); <br/> If (sum [I]-_ max <ans) <br/> ans = sum [I]-_ max; <br/>}< br/> printf ("% d/N", ANS); <br/>}< br/> return 0; <br/>}