HDOJ 5306 Gorgeous Sequence 线段树

http://www.shuizilong.com/house/archives/hdu-5306-gorgeous-sequence/

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 integers n and m denoting the length of the sequence and the number of operations.

The second line contains n separated integers a1,…,an (∀1≤i≤n,0≤ai<231).

Each of the following m lines represents one operation (1≤x≤y≤n,0≤t<231).

It is guaranteed that T=100, ∑n≤1000000, ∑m≤1000000.

 

Output

For every operation of type 1 or 2, print one line containing the answer to the corresponding query.

 

Sample Input

15 51 2 3 4 51 1 52 1 50 3 5 31 1 52 1 5

 

Sample Output

515312
Hint
Please use efficient IO method
 

 

Source

2015 Multi-University Training Contest 2

 

/* ***********************************************Author        :CKbossCreated 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|1void 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);}}/// 清除掉所有大于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);}/// 0void 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);}}/// 1int 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;}/// 2LL 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;}