POJ 3580 SuperMemo

SuperMemo

Time Limit: 5000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3580
64-bit integer IO format: %lld      Java class name: Main

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:

1. 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}
2. 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}
3. 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}
4. 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}
5. DELETE x: delete Ax. 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 {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

 

解题：据说是灰常灰常重口味的splay的题目

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define KT ch[ch[root][1]][0]
using namespace std;
const int maxn = 200010;
const int INF = ~0u>>2;
struct SplayTree{
    int val[maxn],fa[maxn],ch[maxn][2],sz[maxn],root,tot;
    int flip[maxn],add[maxn],minv[maxn],seq[maxn];
    void newnode(int &x,int key,int f){
        x = ++tot;
        minv[x] = val[x] = key;
        ch[x][0] = ch[x][1] = 0;
        sz[x] = 1;
        fa[x] = f;
        add[x] = flip[x] = 0;
    }
    inline void pushdown(int x){
        if(!x) return;
        if(flip[x]){
            flip[ch[x][0]] ^= 1;
            flip[ch[x][1]] ^= 1;
            swap(ch[x][0],ch[x][1]);
            flip[x] = 0;
        }
        if(add[x]){
            if(ch[x][0]){
                add[ch[x][0]] += add[x];
                minv[ch[x][0]] += add[x];
                val[ch[x][0]] += add[x];
            }
            if(ch[x][1]){
                add[ch[x][1]] += add[x];
                minv[ch[x][1]] += add[x];
                val[ch[x][1]] += add[x];
            }
            add[x] = 0;
        }
    }
    inline void pushup(int x){
        minv[x] = min(val[x],min(minv[ch[x][0]],minv[ch[x][1]]));
        sz[x] = 1 + sz[ch[x][0]] + sz[ch[x][1]];
    }
    void rotate(int x,int kd){
        int y = fa[x];
        pushdown(y);
        pushdown(x);
        ch[y][kd^1] = ch[x][kd];
        fa[ch[x][kd]] = y;
        fa[x] = fa[y];
        ch[x][kd] = y;
        fa[y] = x;
        if(fa[x]) ch[fa[x]][y == ch[fa[x]][1]] = x;
        pushup(y);
    }
    void splay(int x,int goal){
        pushdown(x);
        while(fa[x] != goal){
            pushdown(fa[fa[x]]);
            pushdown(fa[x]);
            pushdown(x);
            if(fa[fa[x]] == goal) rotate(x,x == ch[fa[x]][0]);
            else{
                int y = fa[x],z = fa[y],s = (ch[z][0] == y);
                if(x == ch[y][s]){
                    rotate(x,s^1);
                    rotate(x,s);
                }else{
                    rotate(y,s);
                    rotate(x,s);
                }
            }
        }
        pushup(x);
        if(!goal) root = x;
    }
    void build(int &x,int L,int R,int f){
        if(L > R) return;
        int mid = (L + R)>>1;
        newnode(x,seq[mid],f);
        build(ch[x][0],L,mid-1,x);
        build(ch[x][1],mid + 1,R,x);
        pushup(x);
    }
    void init(int n){
        val[0] = root = tot = 0;
        fa[0] = ch[0][0] = ch[0][1] = 0;
        flip[0] = add[0] = sz[0] = 0;
        minv[0] = INF;
        newnode(root,INF,0);
        newnode(ch[root][1],INF,root);
        sz[root] = 2;
        for(int i = 1; i <= n; ++i)
            scanf("%d",seq + i);
        build(KT,1,n,ch[root][1]);
        pushup(ch[root][1]);
        pushup(root);
    }
    void select(int k,int goal){
        int x = root;
        pushdown(x);
        while(sz[ch[x][0]] + 1 != k){
            if(k < sz[ch[x][0]] + 1) x = ch[x][0];
            else{
                k -= sz[ch[x][0]] + 1;
                x = ch[x][1];
            }
            pushdown(x);
        }
        splay(x,goal);
    }
    void MIN(){
        int x,y;
        scanf("%d%d",&x,&y);
        if(x > y) swap(x,y);
        select(x - 1 + 1,0);
        select(y + 1 + 1,root);
        printf("%d
",minv[KT]);
    }
    void Add(){
        int x,y,d;
        scanf("%d%d%d",&x,&y,&d);
        if(x > y) swap(x,y);
        select(x - 1 + 1,0);
        select(y + 1 + 1,root);
        add[KT] += d;
        val[KT] += d;
        minv[KT] += d;
    }
    void reverse(){
        int x,y;
        scanf("%d%d",&x,&y);
        if(x > y) swap(x,y);
        select(x - 1 + 1,0);
        select(y + 1 + 1,root);
        flip[KT] ^= 1;
    }
    void insert(){
        int x,p;
        scanf("%d%d",&x,&p);
        select(x + 1,0);
        select(x + 2,root);
        newnode(KT,p,ch[root][1]);
        pushup(ch[root][1]);
        pushup(root);
    }
    void remove(){
        int x;
        scanf("%d",&x);
        select(x - 1 + 1,0);
        select(x + 1 + 1,root);
        KT = 0;
        pushup(ch[root][1]);
        pushup(root);
    }
    void revolve(){
        int x,y,t;
        scanf("%d%d%d",&x,&y,&t);
        if(x > y) swap(x,y);
        t %= (y - x + 1);
        t = (t + y - x + 1)%(y - x + 1);
        int l = y - t + 1,r = y;
        if(!t) return;
        select(l - 1 + 1,0);
        select(r + 1 + 1,root);
        int tmp = KT;
        KT = 0;
        pushup(ch[root][1]);
        pushup(root);
        select(x - 1 + 1,0);
        select(x + 1,root);//右子树现在是x开始的区间了
        KT = tmp;
        fa[KT] = ch[root][1];
        pushup(ch[root][1]);
        pushup(root);
    }
}spt;
int main(){
    int n,m;
    char op[10];
    scanf("%d",&n);
    spt.init(n);
    scanf("%d",&m);
    while(m--){
        scanf("%s",op);
        if(op[0] == 'A') spt.Add();
        else if(op[0] == 'M') spt.MIN();
        else if(op[0] == 'I') spt.insert();
        else if(op[0] == 'D') spt.remove();
        else if(op[0] == 'R' && op[3] == 'E') spt.reverse();
        else if(op[0] == 'R' && op[3] == 'O') spt.revolve();
    }
    return 0;
}