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HDOJ 5306 Gorgeous Sequence 线段树
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Gorgeous Sequence
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 440 Accepted Submission(s): 87
Problem Description
There is a sequence a of length n . We use ai to denote the i -th element in this sequence. You should do the following three types of operations to this sequence.
0 x y t : For every x≤i≤y , we use min(ai,t) to replace the original ai ‘s value.
1 x y : Print the maximum value of ai that x≤i≤y .
2 x y : Print the sum of ai that x≤i≤y .
Input
The first line of the input is a single integer T , indicating the number of testcases.
The first line contains two integersn and m denoting the length of the sequence and the number of operations.
The second line containsn separated integers a1,…,an (?1≤i≤n,0≤ai<231 ).
Each of the followingm lines represents one operation (1≤x≤y≤n,0≤t<231 ).
It is guaranteed thatT=100 , ∑n≤1000000, ∑m≤1000000 .
The first line contains two integers
The second line contains
Each of the following
It is guaranteed that
Output
For every operation of type 1 or 2 , print one line containing the answer to the corresponding query.
Sample Input
1 5 5 1 2 3 4 5 1 1 5 2 1 5 0 3 5 3 1 1 5 2 1 5
Sample Output
5 15 3 12HintPlease use efficient IO method
Source
field=problem&key=2015+Multi-University+Training+Contest+2&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">2015 Multi-University Training Contest 2
/* *********************************************** Author :CKboss Created Time :2015年07月27日 星期一 08时13分39秒 File Name :HDOJ5306.cpp ************************************************ */ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <string> #include <cmath> #include <cstdlib> #include <vector> #include <queue> #include <set> #include <map> using namespace std; typedef long long int LL; const int maxn=1000030; int n,m; int a[maxn]; struct Node { LL ss; int mx,tag,cv; void toString() { printf("ss: %lld mx: %d tag: %d cv: %d\n",ss,mx,tag,cv); } }T[maxn<<2]; #define lrt (rt<<1) #define rrt (rt<<1|1) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 void push_up(int rt) { T[rt].ss=T[lrt].ss+T[rrt].ss; T[rt].mx=max(T[lrt].mx,T[rrt].mx); T[rt].cv=T[lrt].cv+T[rrt].cv; } void pnc(int t,int l,int r,int rt) { if(T[rt].tag!=0&&T[rt].tag<=t) return ; int all=r-l+1; if(T[rt].cv!=all) { T[rt].ss+=(LL)t*(all-T[rt].cv); T[rt].tag=T[rt].mx=t; T[rt].cv=all; } } void push_down(int l,int r,int rt) { if(T[rt].tag) { int m=(l+r)/2; pnc(T[rt].tag,lson); pnc(T[rt].tag,rson); } } /// 清除掉全部大于t的标记 void fix(int t,int l,int r,int rt) { if(T[rt].mx<t) return ; if(T[rt].tag>=t) T[rt].tag=0; if(l==r) { T[rt].ss=T[rt].mx=T[rt].tag; T[rt].cv=T[rt].tag!=0; } else { push_down(l,r,rt); int m=(l+r)/2; fix(t,lson); fix(t,rson); push_up(rt); } } void build(int l,int r,int rt) { if(l==r) { T[rt].ss=T[rt].mx=T[rt].tag=a[l]; T[rt].cv=1; return ; } T[rt].tag=0; int m=(l+r)/2; build(lson); build(rson); push_up(rt); } /// 0 void update(int L,int R,int t,int l,int r,int rt) { if(T[rt].mx<=t) return ; if(L<=l&&r<=R) { fix(t,l,r,rt); if(l==r) { T[rt].ss=T[rt].mx=T[rt].tag=t; T[rt].cv=1; } else push_up(rt); pnc(t,l,r,rt); } else { push_down(l,r,rt); int m=(l+r)/2; if(L<=m) update(L,R,t,lson); if(R>m) update(L,R,t,rson); push_up(rt); } } /// 1 int query_max(int L,int R,int l,int r,int rt) { if(L<=l&&r<=R) return T[rt].mx; push_down(l,r,rt); int m=(l+r)/2; int ret=0; if(L<=m) ret=max(ret,query_max(L,R,lson)); if(R>m) ret=max(ret,query_max(L,R,rson)); return ret; } /// 2 LL query_sum(int L,int R,int l,int r,int rt) { if(L<=l&&r<=R) return T[rt].ss; push_down(l,r,rt); int m=(l+r)/2; LL ret=0; if(L<=m) ret+=query_sum(L,R,lson); if(R>m) ret+=query_sum(L,R,rson); return ret; } void show(int l,int r,int rt) { printf("rt: %d %d <---> %d\n ",rt,l,r); T[rt].toString(); if(l==r) return ; int m=(l+r)/2; show(lson); show(rson); } /********** fast read **************/ char *ch,buf[40*1024000+5]; void nextInt(int& x) { x=0; for(++ch;*ch<=32;++ch); for(x=0;'0'<=*ch;ch++) x=x*10+*ch-'0'; } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); ch=buf-1; fread(buf,1,1000*35*1024,stdin); int T_T; nextInt(T_T); while(T_T--) { nextInt(n); nextInt(m); for(int i=1;i<=n;i++) nextInt(a[i]); build(1,n,1); int k,l,r,t; while(m--) { nextInt(k); if(k==0) { nextInt(l); nextInt(r); nextInt(t); update(l,r,t,1,n,1); } else if(k==1) { nextInt(l); nextInt(r); printf("%d\n",query_max(l,r,1,n,1)); } else if(k==2) { nextInt(l); nextInt(r); //printf("%lld\n",query_sum(l,r,1,n,1)); printf("%I64d\n",query_sum(l,r,1,n,1)); } } } return 0; }
HDOJ 5306 Gorgeous Sequence 线段树
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