HDU - 3487 Play with Chain __ Splay
2017-03-17 23:24
369 查看
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?
InputThere 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.
OutputFor 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 2 CUT 3 5 4 FLIP 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!
------from ProLights
相关文章推荐
- HDU 3487(Play with Chain-Splay)[template:Splay]
- HDU 3487 Play with Chain(Splay 经典操作)
- 【HDU】3487 Play with Chain Splay
- hdu 3487 Play with Chain(Splay)
- hdu 3487 Play with Chain (Splay树) 区间切割 插入 翻转
- hdu 3487 Play with Chain splay tree
- hdu 3487 Play with Chain(splay区间剪切,翻转)
- HDU 3487 Play with Chain(Splay)
- hdu 3487 Play with Chain splay tree
- Play with Chain HDU - 3487 splay区间翻转,区间位移
- HDU 3487-Play with Chain-splay
- hdu 3487 Play with Chain(splay)
- hdu 3487 Play with Chain (Splay Tree easy)
- HDU 3487 Play with Chain(Splay)
- HDU-3487 Play with Chain Splay tee区间反转,移动
- Splay 模板题 一 hdu 3487 play with chain
- 【HDU 3487】Play with Chain(Splay)
- HDU 3487 Play with Chain | Splay
- HDU-3487 Play with Chain (splay好题 带翻转 cut-link操作)
- HDU 3487 Play with Chain (Splay)