[POJ3468] A Simple Problem with Integers (Treap)

题目链接：http://poj.org/problem?id=3468

这题是线段树的题，拿来学习treap。

不旋转的treap。

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <map>
 #include <set>
 #include <string>
 #include <bitset>
 #include <cmath>
 #include <numeric>
 #include <iterator>
 #include <iostream>
 #include <cstdlib>
 #include <functional>
 #include <queue>
 #include <stack>
 #include <list>
 #include <ctime>
 using namespace std;

 const int inf = ~0U >> ;

 typedef long long LL;

 struct TreapNode{
     int id,sz,key;
     LL val,sum,delta;
     TreapNode* ch[];
     TreapNode(): key(rand()),sz(),id(),val(),sum(),delta() {ch[]=ch[]=NULL;}
     void rz(){
         sz = ;
         sum = val;
         if(ch[]) { sz+=ch[]->sz; sum+=ch[]->sum; }
         if(ch[]) { sz+=ch[]->sz; sum+=ch[]->sum; }
     }
     void pd(){
         if(delta){
             if(ch[]) {
                 ch[]->delta+=delta;
                 ch[]->val+=delta;
                 ch[]->sum+=delta*ch[]->sz;
             }
             if(ch[]) {
                 ch[]->delta+=delta;
                 ch[]->val+=delta;
                 ch[]->sum+=delta*ch[]->sz;
             }
             delta = ;
         }
     }
     ~TreapNode(){
         if(ch[]) delete ch[];
         if(ch[]) delete ch[];
     }
 };

 typedef pair<TreapNode*,TreapNode*> DTreap;

 TreapNode* Merge(TreapNode* A,TreapNode* B) {
     if(!A) return B;
     if(!B) return A;
     A->pd();B->pd();
     if(A->key<B->key) {
         A->ch[] = Merge(A->ch[],B);
         A->rz();
         return A;
     } else {
         B->ch[] = Merge(A,B->ch[]);
         B->rz();
         return B;
     }
 }

 DTreap Split(TreapNode* rt,int id) {
     if(!rt) return DTreap(NULL,NULL);
     DTreap ret;
     rt->pd();
     if(rt->id>id) {
         ret = Split(rt->ch[],id);
         rt->ch[] = ret.second;
         ret.second = rt;
         rt->rz();
         return ret;
     } else {
         ret = Split(rt->ch[],id);
         rt->ch[] = ret.first;
         ret.first = rt;
         rt->rz();
         return ret;
     }
 }

 void Insert(TreapNode*& rt,int id,LL val) {
     DTreap dt = Split(rt,id-);
     TreapNode* lch = dt.first;
     TreapNode* rch = dt.second;
     TreapNode* nd = new TreapNode;
     nd->id = id;
     nd->val = val;
     nd->sum = val;
     lch = Merge(lch,nd);
     rt = Merge(lch,rch);
 }

 void AddInterval(TreapNode*& rt,int l,int r,LL val) {
     if(!rt) return;
     DTreap dt = Split(rt,l-);
     TreapNode* lch = dt.first;
     dt = Split(dt.second,r);
     TreapNode* mid = dt.first;
     TreapNode* rch = dt.second;
     if(mid){
         mid->val += val;
         mid->sum += val*mid->sz;
         mid->delta += val;
     }
     lch = Merge(lch,mid);
     rt = Merge(lch,rch);
 }

 LL QueryInterval(TreapNode*& rt,int l,int r) {
     if(!rt) return ;
     LL ret = ;
     DTreap dt = Split(rt,l-);
     TreapNode* lch = dt.first;
     dt = Split(dt.second,r);
     TreapNode* mid = dt.first;
     TreapNode* rch = dt.second;
     if(mid) ret = mid->sum;
     lch = Merge(lch,mid);
     rt = Merge(lch,rch);
     return ret;
 }

 int n,m;
 LL val;
 char cmd[];

 int main() {
     srand(time(NULL));
     while(~scanf("%d%d",&n,&m)) {
         TreapNode* root = NULL;
         for(int i =  ; i < n ; i++ ) {
             scanf("%lld",&val);
             Insert(root,i+,val);
         }
         for(int i =  ; i < m ; i++ ) {
             scanf("%s",cmd);
             if(cmd[] == 'Q' ) {
                 int l,r;
                 scanf("%d%d",&l,&r);
                 printf("%lld\n",QueryInterval(root,l,r));
             } else {
                 int l,r;
                 LL val;
                 scanf("%d%d%lld",&l,&r,&val);
                 AddInterval(root,l,r,val);
             }
         }
         if(root) delete root;
     }
     return ;
 }