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For a sequence S1,S2,?,SN, and a pair of integers (i,j), if 1≤i≤j≤N and Si<Si+1<Si+2<?<Sj−1<Sj, then the sequence Si,Si+1,?,Sj is a CIS
(Continuous Increasing Subsequence). The longest CIS
of a sequence is called the LCIS
(Longest Continuous Increasing Subsequence).
In this problem, we will give you a sequence first, and then some add
operations and some query
operations. An add operation adds a value to each member in a specified interval. For a query operation, you should output the length of the LCIS of a specified interval.
题目大意就是说求一段区间的最大上升子序列(注意这里的子序列必须是连续的。)
对线段树维护前缀最大上升子序列,后缀最大上升子序列,最大上升子序列,以及区间左端点的值,右端点的值。
这里要注意如果pushUP时,左子树前缀为其长度,那么父亲的前缀的长度应该为左的加右的,右子树同理。
代码如下:
#include<iostream> #include<cstdio> #define max(a,b) (a>b?a:b) #define min(a,b) (a<b?a:b) #define lson L,M,po*2 #define rson M+1,R,po*2+1 using namespace std; struct state { int lef,rig,mid; int x,y; }; int N,Q; state BIT[100005*4]; int COL[100005*4]; int num[100005]; void pushDown(int po) { if(COL[po]) { COL[po*2]+=COL[po]; COL[po*2+1]+=COL[po]; BIT[po*2].x+=COL[po]; BIT[po*2].y+=COL[po]; BIT[po*2+1].x+=COL[po]; BIT[po*2+1].y+=COL[po]; COL[po]=0; } } void pushUP(int L,int R,int po) { BIT[po].mid=max(BIT[po*2].mid,BIT[po*2+1].mid); BIT[po].lef=BIT[po*2].lef; BIT[po].rig=BIT[po*2+1].rig; BIT[po].x=BIT[po*2].x; BIT[po].y=BIT[po*2+1].y; int M=(L+R)/2; if(BIT[po*2].y<BIT[po*2+1].x) { BIT[po].mid=max(BIT[po].mid,BIT[po*2].rig+BIT[po*2+1].lef); if(BIT[po*2].lef==M-L+1) BIT[po].lef=M-L+1+BIT[po*2+1].lef; if(BIT[po*2+1].rig==R-M) BIT[po].rig=R-M+BIT[po*2].rig; } } void build_tree(int L,int R,int po) { if(L==R) { BIT[po].lef=BIT[po].rig=BIT[po].mid=1; scanf("%d",&COL[po]); BIT[po].x=BIT[po].y=COL[po]; return; } int M=(L+R)/2; COL[po]=0; build_tree(lson); build_tree(rson); pushUP(L,R,po); } void update(int ul,int ur,int add,int L,int R,int po) { if(ul<=L&&ur>=R) { COL[po]+=add; BIT[po].x+=add; BIT[po].y+=add; return; } pushDown(po); int M=(L+R)/2; if(ul<=M) update(ul,ur,add,lson); if(ur>M) update(ul,ur,add,rson); pushUP(L,R,po); } int query(int ql,int qr,int L,int R,int po) { if(ql<=L&&qr>=R) return BIT[po].mid; pushDown(po); int M=(L+R)/2; int maxn; if(ql>M) return query(ql,qr,rson); else if(qr<=M) return query(ql,qr,lson); else { maxn=max(query(ql,qr,lson),query(ql,qr,rson)); if(BIT[po*2].y<BIT[po*2+1].x) maxn=max(maxn,min(M-ql+1,BIT[po*2].rig)+min(qr-M,BIT[po*2+1].lef)); return maxn; } } int main() { int T; char ch[5]; int a,b,c; cin>>T; for(int cas=1;cas<=T;++cas) { printf("Case #%d:\n",cas); scanf("%d %d",&N,&Q); build_tree(1,N,1); for(int i=1;i<=Q;++i) { scanf("%s",ch); if(ch[0]==‘q‘) { scanf("%d %d",&a,&b); printf("%d\n",query(a,b,1,N,1)); } else { scanf("%d %d %d",&a,&b,&c); update(a,b,c,1,N,1); } } } return 0; }
(中等) UESTC 360 Another LCIS ,线段树+区间更新。
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原文地址:http://www.cnblogs.com/whywhy/p/4198027.html