Sicily 1802 atomic nucleus US investigation (line segment tree)

The question of the newbie competition reminds me of the time for soy sauce.

// Line segment tree <br/> # include <iostream> <br/> # include <cstdio> <br/> # include <cstring> <br/> # define INF 100000000 <br/> using namespace STD; <br/> int T, M; <br/> char cmd [10], n; <br/> bool R [100005], G [100005]; <br/> struct seg <br/> {<br/> int L, R, _ max, _ min, Delta; // records the maximum value, minimum value, and optimal difference value, _ max, _ Min is 0, indicating that this number does not exist <br/>} tree [100005*4]; <br/> void Update (int x) // update by backtracking <br/>{< br/> tree [X]. _ max = max (tree [2 * X]. _ max, tree [2 * x + 1]. _ max); <B R/> If (! Tree [2 * X]. _ min) tree [X]. _ min = tree [2 * x + 1]. _ min; // If the left subtree does not have a minimum value, use the minimum value of the right subtree <br/> else tree [X]. _ min = tree [2 * X]. _ min; // otherwise, the minimum value of the Left subtree is removed. <br/> tree [X]. delta = min (tree [2 * X]. delta, tree [2 * x + 1]. delta); // obtain the minimum difference value of the optimal subtree <br/> If (tree [2 * X]. _ max & tree [2 * x + 1]. _ min) // if the maximum value of the Left subtree and the minimum value of the right subtree exist <br/> tree [X]. delta = min (tree [X]. delta, tree [2 * x + 1]. _ Min-tree [2 * X]. _ max); <br/>}< br/> void buildtree (int x, int L, int R) <br/>{< br/> tree [x ]. L = L; <br/> tree [X]. R = r; <br/> tree [X]. _ max = 0; <br/> tree [X]. _ min = 0; <br/> tree [X]. delta = inf; <br/> If (L = r) return; <br/> int mid = (L + r)> 1; <br/> buildtree (2 * X, L, mid); <br/> buildtree (2 * x + 1, Mid + 1, R ); <br/>}< br/> void insert (int x, int L, int R) <br/>{< br/> int mid = (tree [X]. L + tree [X]. r)> 1; <br/> If (tree [X]. L = L & tree [X]. R = r) <br/>{< br/> tree [X]. _ max = tree [X]. _ min = L; <br/> retu Rn; <br/>}< br/> If (r <= mid) insert (2 * X, L, R); <br/> else if (L> mid) insert (2 * x + 1, L, R); <br/> tree [X]. _ max = max (tree [2 * X]. _ max, tree [2 * x + 1]. _ max); <br/> If (! Tree [2 * X]. _ min) tree [X]. _ min = tree [2 * x + 1]. _ min; <br/> else tree [X]. _ min = tree [2 * X]. _ min; <br/> Update (x); <br/>}< br/> void remove (int x, int L, int R) <br/> {<br/> int mid = (tree [X]. L + tree [X]. r)> 1; <br/> If (tree [X]. L = L & tree [X]. R = r) <br/>{< br/> tree [X]. _ max = tree [X]. _ min = 0; <br/> tree [X]. delta = inf; <br/> return; <br/>}< br/> If (r <= mid) Remove (2 * X, L, R ); <br/> else if (L> mid) Remove (2 * x + 1, L, R); <br/> Update (x); <br/>}< br/> int query (int x) // directly query the root node to obtain the result <br/>{< br/> If (tree [X]. delta = inf) <br/> return-1; <br/> else return tree [X]. delta; <br/>}</P> <p> int main () <br/>{< br/> int X; <br/> bool first = 1; <br/> // freopen ("in.txt", "r", stdin); <br/> scanf ("% d", & T ); <br/> while (t --) <br/>{< br/> If (first) First = 0; <br/> else printf ("/N "); <br/> buildtree (1,1, 100000); <br/> memset (R, 0, sizeof (R ));/ /Mark whether to remove <br/> memset (G, 0, sizeof (g); // mark whether to generate <br/> scanf ("% d ", & M); <br/> while (M --) <br/>{< br/> scanf ("% s", CMD ); <br/> If (CMD [0] = 'G') <br/> {<br/> scanf ("% d", & X ); <br/> If (! G [x]) <br/>{< br/> G [x] = 1; <br/> insert (1, x, x ); <br/> r [x] = 0; <br/>}< br/> If (CMD [0] = 'R ') <br/>{< br/> scanf ("% d", & X); <br/> If (! R [x]) <br/>{< br/> r [x] = 1; <br/> remove (1, x, x ); <br/> G [x] = 0; <br/>}< br/> If (CMD [0] = 'q ') printf ("% d/N", query (1); <br/>}< br/>}