POJ 3580 SuperMemo

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, ... A_n}. Then the host performs a series of operations and queries on the sequence which consists:

1. ADD x y D: Add D to each number in sub-sequence {A_x ... A_y}. For example, performing "ADD 2 4 1" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5, 5}
2. REVERSE x y: reverse the sub-sequence {A_x ... A_y}. For example, performing "REVERSE 2 4" on {1, 2, 3, 4, 5} results in {1, 4, 3, 2, 5}
3. REVOLVE x y T: rotate sub-sequence {A_x ... A_y} T times. For example, performing "REVOLVE 2 4 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 2, 5}
4. INSERT x P: insert P after A_x. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
5. DELETE x: delete A_x. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
6. MIN x y: query the participant what is the minimum number in sub-sequence {A_x ... A_y}. 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

51 2 3 4 52ADD 2 4 1MIN 4 5

Sample Output

5

Source

POJ Founder Monthly Contest – 2008.04.13, Yao Jinyu

这就是一道splay模板题，不过把各种操作综合在一起时，还是很麻烦啦。

比如说加法标记，就要像线段树一样，先更新了自己，再下传给儿子，更新他们。

虽说这样，还是拍了一晚上，qwqqq

#include<algorithm>#include<iostream>#include<cstdio>using namespace std;const int N=200005;const int inf=1e9+7;int n,m,rt,cnt,a[N],val[N],ch[N][2],fa[N],sz[N],mn[N],add[N],rev[N];void pushup(int x)//¸üÐÂsize {sz[x]=sz[ch[x][0]]+sz[ch[x][1]]+1;mn[x]=val[x];if(ch[x][0])mn[x]=min(mn[x],mn[ch[x][0]]);if(ch[x][1])mn[x]=min(mn[x],mn[ch[x][1]]);}void pushdown(int x)//±ê¼ÇÏ´« {if(add[x]){if(ch[x][0])add[ch[x][0]]+=add[x],mn[ch[x][0]]+=add[x],val[ch[x][0]]+=add[x];if(ch[x][1])add[ch[x][1]]+=add[x],mn[ch[x][1]]+=add[x],val[ch[x][1]]+=add[x];add[x]=0;}if(rev[x]){rev[ch[x][0]]^=1,rev[ch[x][1]]^=1;swap(ch[x][0],ch[x][1]);rev[x]=0;}}int build(int l,int r,int f)//½¨Ê÷ {if(l>r)return 0;int mid=(l+r)/2;int x=++cnt;val[x]=a[mid];ch[x][0]=build(l,mid-1,x);ch[x][1]=build(mid+1,r,x);fa[x]=f;pushup(x);return x;}void rotate(int x,int &k){int y=fa[x],z=fa[y],lc=(ch[y][1]==x),rc=(lc^1);if(y==k)k=x;elsech[z][ch[z][1]==y]=x;fa[x]=z,fa[y]=x,fa[ch[x][rc]]=y;ch[y][lc]=ch[x][rc],ch[x][rc]=y;pushup(y),pushup(x);}void splay(int x,int &k)//½«xÐýתµ½k {while(x!=k){int y=fa[x],z=fa[y];if(y!=k){if((ch[y][0]==x)^(ch[z][0]==y))rotate(x,k);elserotate(y,k);}rotate(x,k);}}int fnd(int x,int rnk)//ÒÑÖªÅÅÃû£¬ÕÒ±àºÅ {pushdown(x);int lc=ch[x][0],rc=ch[x][1];if(rnk<=sz[lc])return fnd(lc,rnk);if(rnk>sz[lc]+1)return fnd(rc,rnk-sz[lc]-1);return x;}void split(int x,int y)//·ÖÀëÇø¼ä[x,y] {splay(x,rt),splay(y,ch[rt][1]);}int main(){scanf("%d",&n);for(int i=2;i<=n+1;i++)scanf("%d",&a[i]);a[1]=a[n+2]=inf;rt=build(1,n+2,0);int x,y,z;char s[11];scanf("%d",&m);while(m--){scanf("%s",s);if(s[0]=='A')//Çø¼ä¼Ó {scanf("%d%d%d",&x,&y,&z);int p=fnd(rt,x),q=fnd(rt,y+2);split(p,q);add[ch[q][0]]+=z,mn[ch[q][0]]+=z,val[ch[q][0]]+=z;}if(s[0]=='R'&&s[3]=='E')//Çø¼ä·´×ª {scanf("%d%d",&x,&y);int p=fnd(rt,x),q=fnd(rt,y+2);split(p,q);rev[ch[q][0]]^=1;}if(s[0]=='R'&&s[3]=='O')//Çø¼äÓÒÒÆ {scanf("%d%d%d",&x,&y,&z);z%=(y-x+1);int nr=y-z,sz=nr-x+1;//ÐÂÇø¼äµÄÓҶ˵ãint p=fnd(rt,x),q=fnd(rt,nr+2);split(p,q);int t=ch[q][0];ch[q][0]=0;pushup(q),pushup(p);p=fnd(rt,y+1-sz),q=fnd(rt,y+2-sz);split(p,q);ch[q][0]=t,fa[t]=q;pushup(q),pushup(p);}if(s[0]=='I')//²åÈë {scanf("%d%d",&x,&y);int p=fnd(rt,x+1),q=fnd(rt,x+2);split(p,q);z=++cnt;val[z]=mn[z]=y;fa[z]=q;sz[z]=1;ch[q][0]=z;pushup(q),pushup(p);}if(s[0]=='D')//ɾ³ý {scanf("%d",&x);int p=fnd(rt,x),q=fnd(rt,x+2);split(p,q);ch[q][0]=0;pushup(q),pushup(p);}if(s[0]=='M')//²éѯ×îСֵ {scanf("%d%d",&x,&y);int p=fnd(rt,x),q=fnd(rt,y+2);split(p,q);printf("%d\n",mn[ch[q][0]]);}}return 0;}