SuperMemo
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 17538 | Accepted: 5496 | |
| Case Time Limit: 2000MS | ||
Description
Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A1, A2, … An}. Then the host performs a series of operations and queries on the sequence which consists:
- ADD x y D: Add D to each number in sub-sequence {Ax … Ay}. For example, performing "ADD 2 4 1" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5, 5}
- REVERSE x y: reverse the sub-sequence {Ax … Ay}. For example, performing "REVERSE 2 4" on {1, 2, 3, 4, 5} results in {1, 4, 3, 2, 5}
- REVOLVE x y T: rotate sub-sequence {Ax … Ay} T times. For example, performing "REVOLVE 2 4 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 2, 5}
- INSERT x P: insert P after Ax. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
- DELETE x: delete Ax. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
- MIN x y: query the participant what is the minimum number in sub-sequence {Ax … Ay}. For example, the correct answer to "MIN 2 4" on {1, 2, 3, 4, 5} is 2
To make the show more interesting, the participant is granted a chance to turn to someone else that means when Jackson feels difficult in answering a query he may call you for help. You task is to watch the TV show and write a program giving the correct answer to each query in order to assist Jackson whenever he calls.
Input
The first line contains n (n ≤ 100000).
The following n lines describe the sequence.
Then follows M (M ≤ 100000), the numbers of operations and queries.
The following M lines describe the operations and queries.
Output
For each "MIN" query, output the correct answer.
Sample Input
5
1
2
3
4
5
2
ADD 2 4 1
MIN 4 5
Sample Output
5
思路&&分析
这题其实直接上无旋Treap就行了。需要注意的是Revolve那个操作中T有可能是负的，或者T%(y-x+1)==0，也就是不动。而且这题没有写数据范围，于是我刚开始#define int long long了，结果T掉了，后来去掉之后稍微改了下inf就过了……
Code
#pragma GCC optimize(3)
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cctype>
#include<climits>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<stack>
#include<climits>
#include<vector>
//#define int long long
using namespace std;
typedef long long ll;
bool Finish_read;
template<class T>inline void read(T &x){Finish_read=0;x=0;int f=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;if(ch==EOF)return;ch=getchar();}while(isdigit(ch))x=x*10+ch-'0',ch=getchar();x*=f;Finish_read=1;}
template<class T>inline void print(T x){if(x/10!=0)print(x/10);putchar(x%10+'0');}
template<class T>inline void writeln(T x){if(x<0)putchar('-');x=abs(x);print(x);putchar('\n');}
template<class T>inline void write(T x){if(x<0)putchar('-');x=abs(x);print(x);}
/*================Header Template==============*/
typedef pair<int,int> Pii;
#define mp make_pair
#define x first
#define y second
const int base=48271,maxn=200005;
int seed=233,cnt,rt;
inline int Rand() {
return seed=(int)(1ll*seed*base%INT_MAX);
}
struct Treap {
int val,sz,ls,rs,tag,rev,fix,mn;
Treap(int _val=0):val(_val){ls=rs=sz=tag=rev=0,fix=Rand();}
}tr[maxn];
inline void Newnode(int val) {
tr[++cnt]=Treap(val);
tr[cnt].sz=1;
}
inline int Sz(int o) {
return o?tr[o].sz:0;
}
inline int Mn(int o) {
return o?tr[o].mn:1073741823;
}
inline int Tag(int o) {
return o?tr[o].tag:1073741823;
}
inline void Pushup(int o) {
tr[o].sz=Sz(tr[o].ls)+1+Sz(tr[o].rs);
tr[o].mn=min(tr[o].val,min(Mn(tr[o].ls)+Tag(tr[o].ls),Mn(tr[o].rs)+Tag(tr[o].rs)));
}
inline void Pushdown(int o) {
if(tr[o].rev) {
if(tr[o].ls)
tr[tr[o].ls].rev^=1;
if(tr[o].rs)
tr[tr[o].rs].rev^=1;
swap(tr[o].ls,tr[o].rs);
tr[o].rev=0;
}
if(tr[o].tag) {
tr[o].val+=tr[o].tag;
tr[o].mn+=tr[o].tag;
if(tr[o].ls)
tr[tr[o].ls].tag+=tr[o].tag;
if(tr[o].rs)
tr[tr[o].rs].tag+=tr[o].tag;
tr[o].tag=0;
}
Pushup(o);
}
inline int Merge(int a,int b) {
if(!a||!b)
return a|b;
Pushdown(a),Pushdown(b);
if(tr[a].fix<tr[b].fix) {
tr[a].rs=Merge(tr[a].rs,b);
Pushup(a);
return a;
}
else {
tr[b].ls=Merge(a,tr[b].ls);
Pushup(b);
return b;
}
}
inline Pii Split(int o,int k) {
if(!o)
return mp(0,0);
Pushdown(o);
if(Sz(tr[o].ls)==k) {
int pre=tr[o].ls;
tr[o].ls=0;
Pushup(o);
return mp(pre,o);
}
if(Sz(tr[o].ls)+1==k) {
int pre=tr[o].rs;
tr[o].rs=0;
Pushup(o);
return mp(o,pre);
}
if(Sz(tr[o].ls)>k) {
Pii tmp=Split(tr[o].ls,k);
tr[o].ls=tmp.y;
Pushup(o);
return mp(tmp.x,o);
}
Pii tmp=Split(tr[o].rs,k-Sz(tr[o].ls)-1);
tr[o].rs=tmp.x;
Pushup(o);
return mp(o,tmp.y);
}
inline void Add(int o,int l,int r,int d) {
Pii tmp1=Split(o,r),tmp2=Split(tmp1.x,l-1);
if(tmp2.y)
tr[tmp2.y].tag+=d;
rt=Merge(Merge(tmp2.x,tmp2.y),tmp1.y);
}
inline void Reverse(int o,int l,int r) {
Pii tmp1=Split(o,r),tmp2=Split(tmp1.x,l-1);
if(tmp2.y)
tr[tmp2.y].rev^=1;
rt=Merge(Merge(tmp2.x,tmp2.y),tmp1.y);
}
inline void Revolve(int o,int l,int r,int t) {
int k=r-l+1;
t=(t%k+k)%k;
if(t==0)
return;
Pii tmp1=Split(o,r),tmp2=Split(tmp1.x,l-1);
Pii tmp3=Split(tmp2.y,k-t);
rt=Merge(Merge(Merge(tmp2.x,tmp3.y),tmp3.x),tmp1.y);
}
inline void Insert(int o,int pos,int k) {
Pii tmp=Split(o,pos);
Newnode(k);
rt=Merge(Merge(tmp.x,cnt),tmp.y);
}
inline void Delete(int o,int xx) {
Pii tmp1=Split(o,xx),tmp2=Split(tmp1.x,xx-1);
rt=Merge(tmp2.x,tmp1.y);
}
inline void Min(int o,int l,int r) {
Pii tmp1=Split(o,r),tmp2=Split(tmp1.x,l-1);
Pushdown(tmp2.y);
printf("%d\n",tr[tmp2.y].mn);
rt=Merge(Merge(tmp2.x,tmp2.y),tmp1.y);
}
signed main() {
int n,q;
read(n);
for(int i=1;i<=n;i++) {
int x;
read(x);
Insert(rt,i-1,x);
}
read(q);
while(q--) {
char op[10];
scanf("%s",op);
if(op[0]=='A') {
int xx,yy,d;
read(xx),read(yy),read(d);
Add(rt,xx,yy,d);
}
if(op[0]=='R') {
int xx,yy;
read(xx);read(yy);
if(op[3]=='E')
Reverse(rt,xx,yy);
else {
int T;
read(T);
Revolve(rt,xx,yy,T);
}
}
if(op[0]=='I') {
int xx,p;
read(xx),read(p);
Insert(rt,xx,p);
}
if(op[0]=='D') {
int xx;
read(xx);
Delete(rt,xx);
}
if(op[0]=='M') {
int xx,yy;
read(xx),read(yy);
Min(rt,xx,yy);
}
}
}