POJ 3468 A Simple Problem with Integers (線段樹)

題目大意：給一序列，對子序列的每個數進行增量，查詢某子序列的和。

 

思路：建立一線段樹，節點保存區間左右端點，區間插入的和，區間的增量(必須滿足區間吻合)，左右孩子。

 

增量的時候注意在往下一個節點查找的同時，必須把增量累加到大的區間裏，直到查找到區間吻合的。

 

nInc 必須用64位整型，否則WA。

 

查詢的時候把nInc往下帶1600+ms，不往下帶1500+ms，竟然時間更短。······

 

 

 

#include <iostream> using namespace std ; struct Node { int L,R; Node *pLeft,*pRight; __int64 nSum; __int64 nInc; }; Node Tree[270000]; int nCount; __int64 nSum; int Mid(Node *pRoot) { return (pRoot->L+pRoot->R)>>1; } void Build(Node *pRoot,int L,int R) { pRoot->L=L; pRoot->R=R; pRoot->nSum=pRoot->nInc=0; if(L==R) return ; pRoot->pLeft=Tree+nCount; nCount++; pRoot->pRight=Tree+nCount; nCount++; Build(pRoot->pLeft,L,Mid(pRoot)); Build(pRoot->pRight,Mid(pRoot)+1,R); } void Insert(Node *pRoot,int i,int v) { pRoot->nSum+=v; if(pRoot->L==i && pRoot->R==i) return ; if(i<=Mid(pRoot)) Insert(pRoot->pLeft,i,v); else Insert(pRoot->pRight,i,v); } void Increase(Node *pRoot,int L,int R,int v) { if(pRoot->L==L && pRoot->R==R) { pRoot->nInc+=v; return ; } pRoot->nSum+= v*(R-L+1); if(R<=Mid(pRoot)) Increase(pRoot->pLeft,L,R,v); else if(L>Mid(pRoot)) Increase(pRoot->pRight,L,R,v); else { Increase(pRoot->pLeft,L,Mid(pRoot),v); Increase(pRoot->pRight,Mid(pRoot)+1,R,v); } } void Query(Node *pRoot,int L,int R) { nSum+=(R-L+1)*pRoot->nInc; if(pRoot->L==L && pRoot->R==R) { nSum+=pRoot->nSum; //nSum+=pRoot->nSum+pRoot->nInc*(R-L+1); return ; } /* if(pRoot->nInc) { pRoot->nSum+=(pRoot->R-pRoot->L+1)*pRoot->nInc; Increase(pRoot->pLeft,pRoot->L,Mid(pRoot),pRoot->nInc); Increase(pRoot->pRight,Mid(pRoot)+1,pRoot->R,pRoot->nInc); pRoot->nInc=0; } */ if(R<=Mid(pRoot)) Query(pRoot->pLeft,L,R); else if(L>Mid(pRoot)) Query(pRoot->pRight,L,R); else { Query(pRoot->pLeft,L,Mid(pRoot)); Query(pRoot->pRight,Mid(pRoot)+1,R); } } int main() { int N,Q; char cmd; int i,v,L,R; while(scanf("%d%d",&N,&Q)!=EOF) { nCount=1; Build(Tree,1,N); for(i=1;i<=N;i++) { scanf("%d",&v); Insert(Tree,i,v); } while(Q--) { getchar(); scanf("%c",&cmd); if(cmd=='C') { scanf("%d%d%d",&L,&R,&v); Increase(Tree,L,R,v); } else { scanf("%d%d",&L,&R); nSum=0; Query(Tree,L,R); printf("%I64d/n",nSum); } } } return 0; }