HDU

Play with Chain

 HDU - 3487

YaoYao is fond of playing his chains. He has a chain containing n diamonds on it. Diamonds are numbered from 1 to n. 
At first, the diamonds on the chain is a sequence: 1, 2, 3, …, n. 
He will perform two types of operations: 
CUT a b c: He will first cut down the chain from the ath diamond to the bth diamond. And then insert it after the cth diamond on the remaining chain. 
For example, if n=8, the chain is: 1 2 3 4 5 6 7 8; We perform “CUT 3 5 4”, Then we first cut down 3 4 5, and the remaining chain would be: 1 2 6 7 8. Then we insert “3 4 5” into the chain before 5th diamond, the chain turns out to be: 1 2 6 7 3 4 5 8. 

FLIP a b: We first cut down the chain from the ath diamond to the bth diamond. Then reverse the chain and put them back to the original position. 
For example, if we perform “FLIP 2 6” on the chain: 1 2 6 7 3 4 5 8. The chain will turn out to be: 1 4 3 7 6 2 5 8 

He wants to know what the chain looks like after perform m operations. Could you help him? 

Input

There will be multiple test cases in a test data. 
For each test case, the first line contains two numbers: n and m (1≤n, m≤3*100000), indicating the total number of diamonds on the chain and the number of operations respectively. 
Then m lines follow, each line contains one operation. The command is like this: 
CUT a b c // Means a CUT operation, 1 ≤ a ≤ b ≤ n, 0≤ c ≤ n-(b-a+1). 
FLIP a b    // Means a FLIP operation, 1 ≤ a < b ≤ n. 
The input ends up with two negative numbers, which should not be processed as a case. 

Output

For each test case, you should print a line with n numbers. The ith number is the number of the ith diamond on the chain.

Sample Input

8 2CUT 3 5 4FLIP 2 6-1 -1

Sample Output

1 4 3 7 6 2 5 8

Source

HDU - 3487

My Solution

题意：对1~n这n个数，进行m次操作，分别可以进行区间移动和区间反转，求最终的序列。

Splay

Splay的基础题，按照要求进行区间移动和区间反转即可。

复杂度 O(m*logn)

#include <iostream>#include <cstdio>#include <string>#include <algorithm>using namespace std;typedef long long LL;const int MAXN = 3e5 + 8;int n;bool flag;struct SplayTree{    int ch[MAXN][2], fa[MAXN], tot, root;    int val[MAXN], reversed[MAXN], sz[MAXN];    int newnode(int f, int u){        reversed[tot] = ch[tot][0] = ch[tot][1] = 0;        fa[tot] = f; val[tot] = u; sz[tot] = 1;        return tot ++;    }    void init(){        tot = 1; ch[0][0] = ch[0][1] = fa[0] = sz[0] = 0;    }    void build(int f, int &x, int l, int r){        if (l > r) return;        int mid = (l + r) >> 1;        x = newnode(f, mid);        build(x, ch[x][0], l, mid - 1);        build(x, ch[x][1], mid + 1, r);        pushup(x);    }    void pushdown(int x){        if(!reversed[x]) return;        reversed[ch[x][0]] ^= 1;        reversed[ch[x][1]] ^= 1;        swap(ch[x][0], ch[x][1]);        reversed[x] = 0;    }    void pushup(int x){        sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;    }    void _rotate(int x, int kind){        int y = fa[x];        ch[y][!kind] = ch[x][kind];        fa[ch[x][kind]] = y;        if(fa[y]) ch[fa[y]][y == ch[fa[y]][1]] = x;        fa[x] = fa[y]; fa[y] = x;        ch[x][kind] = y;        sz[x] = sz[y];        pushup(y);    }    void splay(int x, int r){        int y, z;        for (int f = fa[r]; fa[x] != f; ){            if(fa[fa[x]] == f){_rotate(x, x == ch[fa[x]][0]); return; }            y = x == ch[fa[x]][0], z = fa[x] == ch[fa[fa[x]]][0];            y^z ? (_rotate(x, y), _rotate(x, z)) : (_rotate(fa[x], z), _rotate(x, y));        }    }    void _find(int &x, int k){        for (int i = x, j = k; ; ){            pushdown(i);            if(sz[ch[i][0]] == j){splay(i, x); x = i; break; }            if (sz[ch[i][0]] > j){i = ch[i][0]; continue; }            j -= sz[ch[i][0]] + 1; i = ch[i][1];        }    }    void _reverse(int&x, int l, int r){        _find(x, l - 1);        _find(ch[x][1], r - l + 1);        reversed[ch[ch[x][1]][0]] ^= 1;    }    void cut_insert(int&x, int l, int r, int c){        _find(x, l - 1);        _find(ch[x][1], r - l + 1);        int add = sz[ch[ch[x][1]][0]], temp = ch[ch[x][1]][0];        sz[x] -= add; sz[ch[x][1]] -= add;        ch[ch[x][1]][0] = 0;        _find(x, c);        _find(ch[x][1], 0);        ch[ch[x][1]][0] = temp;        fa[temp] = ch[x][1];        sz[ch[x][1]] += add;        sz[x] += add;    }    void traversal(int x){        pushdown(x);        if(ch[x][0]) traversal(ch[x][0]);        if(val[x] != 0 && val[x] != n + 1){            if(flag) cout << " ";            cout << val[x];            flag = true;        }        if(ch[x][1]) traversal(ch[x][1]);    }}spt;string s;int main(){   #ifdef LOCAL    freopen("SplayTree.in", "r", stdin);    //freopen("SplayTree.out", "w", stdout);    #endif // LOCAL    ios::sync_with_stdio(false); cin.tie(0);    int m, l, r, c;    while(cin >> n >> m){        if(n == -1) break;        spt.init(); spt.root = 0;        spt.build(0, spt.root, 0, n + 1);        while(m--){            cin >> s;            cin >> l >> r;            if(s[0] == 'C') cin >> c, spt.cut_insert(spt.root, l, r, c);            else spt._reverse(spt.root, l, r);        }        flag = false;        spt.traversal(spt.root);        cout << "\n";    }    return 0;}

Thank you!

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