Poj issue .a simple problem with integers Problem Solving report

Using a tree array and a line segment tree is relatively simple. Here we use this question to learn splay.

The first write, the code is ugly

/* Initialize and add a node with a sufficient key value to ensure that each required node has a successor */# include <iostream> # include <cstdio> # define ll long longusing namespace STD; const int maxn = 1111111, INF = 0x7fffffff; struct node {// The required record information ll key, Val, sum, lazy, size, CNT; // The pointer to the Son and Father: node * Ch [2], * pre; node () {pre = CH [0] = CH [1] = 0; size = 1; key = 0;} node (LL key): Key (key) {pre = CH [0] = CH [1] = 0; size = 1, CNT = 1, lazy = 0;} void csize () {size = CNT; If (CH [0]! = NULL) size + = CH [0]-> size; If (CH [1]! = NULL) size + = CH [1]-> size;} void csum () {sum = val; If (CH [0]! = NULL) sum + = CH [0]-> sum + CH [0]-> size * Ch [0]-> lazy; If (CH [1]! = NULL) sum + = CH [1]-> sum + CH [1]-> size * Ch [1]-> lazy ;}} nil (0 ), * nil = & nil; struct splay {node * root, nod [maxn]; int ncnt; // calculate the number of knots with different key values, de-duplicated splay () {ncnt = 0; root = & (nod [ncnt ++] = node (INF); root-> pre = nil; root-> val = root-> sum = 0;} void push_down (node * X) {If (X-> lazy! = 0) {If (X-> CH [0]! = NULL) x-> CH [0]-> lazy + = x-> lazy; If (X-> CH [1]! = NULL) x-> CH [1]-> lazy + = x-> lazy; X-> Val + = x-> lazy;} X-> lazy = 0 ;} void Update (node * X) {X-> csize (); X-> csum ();} void rotate (node * X, int Sta) {// single rotation operation, 0 left-hand, 1 right-hand node * P = x-> pre, * g = p-> pre; push_down (P), push_down (x); P-> CH [! Sta] = x-> CH [sta]; If (X-> CH [sta]! = NULL) x-> CH [sta]-> pre = P; X-> pre = g; If (G! = Nil) if (G-> CH [0] = P) g-> CH [0] = x; else G-> CH [1] = X; x-> CH [sta] = p, p-> pre = x, update (p); If (P = root) root = x ;} void splay (node * X, node * Y) {// splay operation, which means to convert node X to the root for (push_down (x); X-> pre! = Y;) {// pass the X mark down if (X-> pre = y) {// The target node is the parent node if (X-> pre-> CH [0] = x) rotate (x, 1); else rotate (x, 0 );} else {node * P = x-> pre, * g = p-> pre; If (G-> CH [0] = P) if (p-> CH [0] = x) rotate (P, 1), rotate (x, 1); // One-font double rotation else rotate (X, 0), rotate (x, 1); // <G id = "2", double rotation else if (p-> CH [1] = x) rotate (p, 0 ), rotate (x, 0); // \ one-font rotation else rotate (x, 1), rotate (x, 0); //> glyph rotation} Update (x ); // Maintain the x node} // locate the K node that is convenient in the middle order and rotate it to the bottom of node y. Void select (int K, node * Y) {int TEM; node * t; for (t = root;) {push_down (t ); // mark the downloading TEM = T-> CH [0]-> size; If (k = TEM + 1) break; // locate the k-th node T if (k <= TEM) t = T-> CH [0]; // The k-th node is in the left subtree else K-= TEM + 1, t = T-> CH [1]; // In the right subtree} splay (t, y);} bool search (LL key, node * Y) {node * t = root; For (; t! = NULL;) {push_down (t); If (t-> key & T-> CH [0]! = NULL) t = T-> CH [0]; else if (t-> key <key & T-> CH [1]! = NULL) t = T-> CH [1]; else break;} splay (T, Y); Return T-> key = key;} void insert (INT key, int Val) {If (search (Key, nil) Root-> CNT ++, root-> size ++; else {int d = Key> root-> key; node * t = & (nod [++ ncnt] = node (key); push_down (Root); t-> val = T-> sum = val; t-> CH [d] = root-> CH [d]; If (root-> CH [d]! = NULL) Root-> CH [d]-> pre = T; t-> CH [! D] = root; t-> pre = root-> pre; root-> pre = T; root-> CH [d] = NULL; Update (Root ); root = T;} Update (Root) ;}} sp; ll n, m, X; int main () {scanf ("% LLD", & N, & M); SP. insert (0, 0); SP. insert (n + 1, 0); For (INT I = 1; I <= N; I ++) {scanf ("% LLD", & X); SP. insert (I, x);} Char cmd; int L, R; For (INT I = 1; I <= m; I ++) {scanf ("\ n % C % d", & cmd, & L, & R); SP. search (L-1, nil); SP. search (R + 1, SP. root); If (cmd = 'q') {node * t = sp. root-> CH [1]-> CH [0]; ll ans = T-> sum + T-> size * t-> lazy; printf ("% LLD \ n", ANS);} If (cmd = 'C') {ll C; scanf ("% LLD", & C); SP. root-> CH [1]-> CH [0]-> lazy + = C ;}} return 0 ;}

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Poj issue .a simple problem with integers Problem Solving report