题目：http://poj.org/problem?id=3580
SuperMemo
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 13105 | Accepted: 4104 | |
| 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
Source
POJ Founder Monthly Contest – 2008.04.13, Yao Jinyu
题意：给定一个序列，每次执行一个操作，对于每个min输出即可。
题解：
好个码农题。。。
Splay处理一下区间翻转，区间加上k，区间左右移动(循环序列)，区间查询。。。
记得开long long。。。。。。
代码:
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define MAXN 100010 4 #define MAXM 100010 5 #define INF 1e9 6 #define LL long long 7 struct node 8 { 9 LL left,right,mn,val,size; 10 }tree[MAXN+MAXM]; 11 LL a[MAXN+MAXM],father[MAXN+MAXM],rev[MAXN+MAXM],tag[MAXN+MAXM]; 12 LL read() 13 { 14 LL s=0,fh=1;char ch=getchar(); 15 while(ch<'0'||ch>'9'){if(ch=='-')fh=-1;ch=getchar();} 16 while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');ch=getchar();} 17 return s*fh; 18 } 19 void Pushup(LL x) 20 { 21 LL l=tree[x].left,r=tree[x].right; 22 tree[x].size=tree[l].size+tree[r].size+1; 23 tree[x].mn=min(min(tree[l].mn,tree[r].mn),tree[x].val); 24 } 25 void Build(LL l,LL r,LL f) 26 { 27 if(l>r)return; 28 LL now=l,last=f; 29 if(l==r) 30 { 31 tree[now].val=tree[now].mn=a[l];father[now]=last; 32 tree[now].size=1; 33 if(l<f)tree[last].left=now; 34 else tree[last].right=now; 35 } 36 LL mid=(l+r)/2; 37 now=mid; 38 Build(l,mid-1,mid);Build(mid+1,r,mid); 39 father[now]=last;tree[now].val=a[mid]; 40 Pushup(now); 41 if(mid<f)tree[last].left=now; 42 else tree[last].right=now; 43 } 44 /*void Pushup(int x) 45 { 46 int l=tree[x].left,r=tree[x].right; 47 tree[x].size=tree[l].size+tree[r].size+1; 48 tree[x].mn=min(min(tree[l].mn,tree[r].mn),tree[x].val); 49 }*/ 50 void rotate(LL x,LL &root) 51 { 52 LL y=father[x],z=father[y]; 53 if(y==root)root=x; 54 else 55 { 56 if(tree[z].left==y)tree[z].left=x; 57 else tree[z].right=x; 58 } 59 if(tree[y].left==x) 60 { 61 father[x]=z;father[y]=x;tree[y].left=tree[x].right;tree[x].right=y;father[tree[y].left]=y; 62 } 63 else 64 { 65 father[x]=z;father[y]=x;tree[y].right=tree[x].left;tree[x].left=y;father[tree[y].right]=y; 66 } 67 Pushup(y);Pushup(x); 68 } 69 void Splay(LL x,LL &root) 70 { 71 while(x!=root) 72 { 73 int y=father[x],z=father[y]; 74 if(y!=root) 75 { 76 if((tree[y].left==x)^(tree[z].left==y))rotate(x,root); 77 else rotate(y,root); 78 } 79 rotate(x,root); 80 } 81 } 82 void Pushdown(LL x) 83 { 84 LL l=tree[x].left,r=tree[x].right; 85 if(tag[x]!=0) 86 { 87 tag[l]+=tag[x];tag[r]+=tag[x]; 88 tree[l].val+=tag[x];tree[r].val+=tag[x]; 89 tree[l].mn+=tag[x];tree[r].mn+=tag[x]; 90 tag[x]=0; 91 } 92 if(rev[x]!=0) 93 { 94 rev[l]^=1;rev[r]^=1;rev[x]^=1; 95 swap(tree[x].left,tree[x].right); 96 } 97 } 98 LL Find(LL root,LL rank) 99 { 100 Pushdown(root); 101 if(tree[tree[root].left].size+1==rank)return root; 102 else if(rank<=tree[tree[root].left].size)return Find(tree[root].left,rank); 103 else return Find(tree[root].right,rank-tree[tree[root].left].size-1); 104 } 105 int main() 106 { 107 LL n,m,i,rt,SIZE,l,r,add,x,y,z,L,R,T,X,P; 108 char fh[8]; 109 n=read(); 110 tree[0].val=INF;a[0]=tree[0].mn=INF; 111 tree[1].val=INF;tree[n+2].val=INF; 112 tree[1].mn=INF;tree[n+2].mn=INF; 113 a[1]=INF;a[n+2]=INF; 114 for(i=2;i<=n+1;i++)a[i]=read(),tree[i].val=tree[i].mn=INF; 115 Build(1,n+2,0); 116 SIZE=n+2;rt=(1+n+2)/2; 117 m=read(); 118 for(i=1;i<=m;i++) 119 { 120 scanf(" %s",fh); 121 if(fh[0]=='A') 122 { 123 l=read();r=read();add=read(); 124 x=Find(rt,l);y=Find(rt,r+2); 125 Splay(x,rt);Splay(y,tree[x].right); 126 z=tree[y].left; 127 tag[z]+=add;tree[z].val+=add;tree[z].mn+=add; 128 } 129 else if(fh[0]=='R') 130 { 131 if(fh[3]=='E') 132 { 133 l=read();r=read(); 134 x=Find(rt,l);y=Find(rt,r+2); 135 Splay(x,rt);Splay(y,tree[x].right); 136 z=tree[y].left; 137 rev[z]^=1; 138 } 139 else 140 { 141 l=read();r=read();T=read(); 142 L=l;R=r; 143 T=(T%(r-l+1)+(r-l+1))%(r-l+1); 144 if(T==0)continue; 145 l=r-T+1; 146 x=Find(rt,l);y=Find(rt,r+2); 147 Splay(x,rt);Splay(y,tree[x].right); 148 z=tree[y].left; 149 father[z]=0;tree[y].left=0; 150 Pushup(y);Pushup(x); 151 x=Find(rt,L);y=Find(rt,L+1); 152 Splay(x,rt);Splay(y,tree[x].right); 153 father[z]=y;tree[y].left=z; 154 Pushup(y);Pushup(x); 155 } 156 } 157 else if(fh[0]=='I') 158 { 159 X=read();P=read(); 160 x=Find(rt,X+1);y=Find(rt,X+2); 161 Splay(x,rt);Splay(y,tree[x].right); 162 tree[y].left=++SIZE;tree[SIZE].val=P; 163 father[SIZE]=y;tree[SIZE].size=1; 164 tree[SIZE].mn=P; 165 Pushup(y);Pushup(x); 166 } 167 else if(fh[0]=='D') 168 { 169 X=read(); 170 x=Find(rt,X);y=Find(rt,X+2); 171 Splay(x,rt);Splay(y,tree[x].right); 172 z=tree[y].left;tree[y].left=0; 173 tree[z].size=0;father[z]=0; 174 //tree[SIZE].val=INF;tree[SIZE].mn=INF; 175 Pushup(y);Pushup(x); 176 } 177 else 178 { 179 l=read();r=read(); 180 x=Find(rt,l);y=Find(rt,r+2); 181 Splay(x,rt);Splay(y,tree[x].right); 182 z=tree[y].left; 183 printf("%lld ",tree[z].mn); 184 } 185 } 186 fclose(stdin); 187 fclose(stdout); 188 return 0; 189 }