spoj 1470 Another Sequence Problem

本题是练习伸展树操作的绝好题目，写了3天，悲剧。。。

题目链接：http://www.spoj.pl/problems/SEQ2/

我因为y和n的顺序写错，导致TLE了一天。。再次悲剧。。

重要的地方我都有注释：

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAX = 500000 * 2;const int INF = 1000000000;inline int max(int a,int b){return a>b?a:b;}inline int min(int a,int b){return a<b?a:b;}struct Node{int sum;int value;int size;Node *ch[2];Node * pre;bool rev;bool same;int MaxL;int MaxR;int MaxM;};struct Splay{Node  data[MAX];Node *store[MAX];//人工堆栈Node * null;Node * root;int number;int top;Node * newNode(int val){Node * p;if(top>0){p = store[top--];}else{p = &data[number++];}p->size = 1;p->value = val;p->sum = val;p->rev = p->same = false;p->MaxL = p->MaxR = p->MaxM = val;p->ch[0] = p->ch[1] = null;p->pre = null;return p;}void init(){number = 0;top = 0;null = newNode(-INF);null->size = null->sum = 0;root = newNode(-INF);root->ch[1] = newNode(-INF);root->ch[1]->pre = root;push_up(root);//这里主要是为了更新size，sum的更新并不正确，但是在rotate和splay操作中会正确}void push_up(Node * p){if(p == null){return;}push_down(p);push_down(p->ch[0]);//以求和为例，p节点值的更新需要依赖两个字节点，那么这两个字节点的sum要先得到更新，所以要下放push_down(p->ch[1]);p->size = p->ch[0]->size + p->ch[1]->size + 1;p->sum = p->ch[0]->sum + p->ch[1]->sum + p->value;p->MaxL = max(p->ch[0]->MaxL,p->ch[0]->sum + p->value + max(0,p->ch[1]->MaxL));p->MaxR = max(p->ch[1]->MaxR,p->ch[1]->sum + p->value + max(0,p->ch[0]->MaxR));    p->MaxM=max(p->ch[0]->MaxM,p->ch[1]->MaxM);    p->MaxM=max(p->MaxM,max(p->ch[0]->MaxR+p->ch[1]->MaxL,0)+p->value);    p->MaxM=max(p->MaxM,max(p->ch[0]->MaxR,p->ch[1]->MaxL)+p->value);}void push_down(Node * p){if(p == null){return;}if(p->rev){p->rev = false;p->ch[0]->rev ^=1;p->ch[1]->rev ^=1;swap(p->ch[0],p->ch[1]);swap(p->MaxL,p->MaxR);}if(p->same){p->same = false;p->ch[0]->same = p->ch[1]->same = true;p->ch[0]->value = p->ch[1]->value = p->value;p->sum = p->value * p->size;p->MaxL = p->MaxR = p->MaxM = p->size * p->value;if(p->value <0){p->MaxL = p->MaxR = p->MaxM = p->value;}}}//中序遍历void dfs(Node * p){if(p == null){return;}dfs(p->ch[0]);printf("%d   ",p->value);dfs(p->ch[1]);}//c=1 右旋void rotate(Node * x,int c){Node * y = x->pre;push_down(y);push_down(x);y->ch[!c] = x->ch[c];if(x->ch[c]!=null){x->ch[c]->pre = y;}x->pre = y->pre;if(y->pre !=null){if(y->pre->ch[0] == y){y->pre->ch[0] = x;}else{y->pre->ch[1] = x;}}x->ch[c] = y;y->pre = x;push_up(y);if(y == root){root = x;}}void splay(Node * x,Node * f){if(x == null){return;}for(push_down(x);x->pre!=f;){if(x->pre->pre == f)//单旋{if(x == x->pre->ch[0]){rotate(x,1);}else{rotate(x,0);}}else{Node * y = x->pre;Node * z = y->pre;if(z->ch[0] == y){if(y->ch[0] == x)//一字形旋转{rotate(y,1);rotate(x,1);}else//之字形{rotate(x,0);rotate(x,1);}}else{if(y->ch[1] == x)//一字形{rotate(y,0);rotate(x,0);}else//之字形{rotate(x,1);rotate(x,0);}}}}push_up(x);}void select(int k,Node * f)//k是指第几个节点,从1开始{Node * t;//k = k-1+1;//其实正常情况下应该是 k = k-1;但是前后加两个节点的话，正好向右移动一位 for(t=root;;){push_down(t);int size = t->ch[0]->size;if(k == size){break; }if(k < size){t = t->ch[0];}else{k -= size + 1;t = t->ch[1];}}splay(t,f);}Node * build(int left,int right,int *ary){if(left>right){return null;}int mid = (left + right)>>1;Node * p  =  newNode(ary[mid]);p->ch[0] = build(left,mid-1,ary);if(p->ch[0]!=null){p->ch[0]->pre = p;}p->ch[1] = build(mid+1,right,ary);if(p->ch[1]!=null){p->ch[1]->pre = p;}push_up(p);//Update操作return p;}void insert(int pos,int *ary,int n){select(pos-1,null);select(pos,root);Node * p = build(0,n-1,ary);root->ch[1]->ch[0] = p;p->pre = root->ch[1];splay(p,null);//将p移至根节点}void del(int pos,int end){select(pos-1,null);select(end,root);Node * p = root->ch[1]->ch[0];//如果不回收的话，将会造成内存泄漏root->ch[1]->ch[0] = null;splay(root->ch[1],null);//别忘了维护recyle(p);}void recyle(Node *p)//人工回收删除的空间，防止栈溢出{if(p == null){return;}recyle(p->ch[0]);recyle(p->ch[1]);store[++top] = p;}int getSum(int start,int end){select(start-1,null);select(end,root);//这里完全不用担心，肯定会移到root的右子树，因为end比start-1大，根据查询二叉树的性质return root->ch[1]->ch[0]->sum;}int maxSum(int pos,int end){select(pos-1,null);select(end,root);Node * p = root->ch[1]->ch[0];return p->MaxM;}void makeSame(int pos,int end,int s){select(pos-1,null);select(end,root);Node * p = root->ch[1]->ch[0];p->same = true;p->value = s;splay(p,null);}void reverse(int pos,int end){select(pos-1,null);select(end,root);Node * p = root->ch[1]->ch[0];p->rev ^= 1;splay(p,null);}}sp;int ary[MAX];int main(){int T;#ifndef ONLINE_JUDGEfreopen("in.txt","r",stdin);#endiffor(scanf("%d",&T);T--;){int n,m;int x,y;int s;Node * p;char cmd[20];scanf("%d%d",&n,&m);for(int i=0;i<n;i++){scanf("%d",&ary[i]);}sp.init();sp.insert(1,ary,n); for(;m--;){scanf("%s",cmd);if(strcmp(cmd,"GET-SUM") == 0){scanf("%d%d",&x,&y);printf("%d\n",sp.getSum(x,x+y));}if(strcmp(cmd,"MAX-SUM") == 0){printf("%d\n",sp.maxSum(1,sp.root->size - 1));}if(strcmp(cmd,"INSERT") == 0){scanf("%d%d",&x,&y);x++;//向后移动一位for(int i=0;i<y;i++){scanf("%d",&ary[i]);}sp.insert(x,ary,y);}if(strcmp(cmd,"DELETE") == 0){scanf("%d%d",&x,&y);sp.del(x,x+y);}if(strcmp(cmd,"REVERSE") == 0){scanf("%d%d",&x,&y);sp.reverse(x,x+y);}if(strcmp(cmd,"MAKE-SAME") == 0){scanf("%d%d%d",&x,&y,&s);sp.makeSame(x,x+y,s);}}//dfs(root);//printf("\n");}return 0;}