[BZOJ1014] [JSOI2008]火星人prefix && splay+字符串hash 重写版

最新推荐文章于 2018-09-24 22:28:00 发布

原创最新推荐文章于 2018-09-24 22:28:00 发布 · 521 阅读

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本文回顾了去年在代码实现中遇到的挑战，并分享了如何通过优化数据结构（如使用Hash表）来提高效率的经验。同时指出，尽管代码改进显著，但实际运行速度仍存在问题，暗示了在追求技术创新的同时，也需关注算法复杂度和优化实践的重要性。

看着去年十二月那个5K+的代码我突然觉得过去的我还是蛮拼的

用Hash维护一棵子树的信息更改和询问都比较方便可以参考原来的那篇

对于插入其实应该找到把x-1旋转到根 x+1旋转到根的右儿子然后再插到x+1的做儿子处

感觉这样靠谱些而且用数组版的splay代码少了好几K

但是还是跑了9s 慢的要死啊Orz

果然像我这样的人最好早点滚粗

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define SF scanf
#define PF printf
using namespace std;
typedef long long LL;
const int MAXN = 150000;
const int MOD = 199999;
int base[MAXN+10];
int n, m;
char c;
char s[MAXN+10];
struct Splay_Tree{
	int val[MAXN+10], sz[MAXN+10], fa[MAXN+10];
	int RK[MAXN+10];
	int ncnt, root, ch[MAXN+10][2];
	void up(int x) {
		sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;
		RK[x] = (1LL * RK[ch[x][0]] * base[sz[ch[x][1]]+1] % MOD +
				1LL * val[x] * base[sz[ch[x][1]]] % MOD +
				RK[ch[x][1]]) % MOD;
	}
	void NewNode(int &x, int key, int pre) {
		x = ++ncnt;
		val[x] = key; sz[x] = 1; fa[x] = pre;
		ch[x][0] = ch[x][1] = 0;
	}
	void Rotate(int x) {
		int y = fa[x], z = fa[y];
		bool d = x == ch[y][0];
		if(ch[y][!d] = ch[x][d]) fa[ch[x][d]] = y;
		if(fa[x] = z) ch[z][y == ch[z][1]] = x;
		up(ch[x][d] = y); fa[y] = x;
	}
	void splay(int x, int goal) {
		for(int y = fa[x]; y ^ goal; Rotate(x), y = fa[x])
			if(fa[y] != goal) Rotate((x == ch[y][0]) && (y == ch[fa[y]][0]) ? y : x);
		up(x); if(!goal) root = x;
	}
	int Find_kth(int k) {
		int x = root;
		if(k < 0 || k > sz[x]) return -1;
		while(k <= sz[ch[x][0]] || k > sz[ch[x][0]] + 1) 
			if(k <= sz[ch[x][0]]) x = ch[x][0];
			else k -= sz[ch[x][0]] + 1, x = ch[x][1];
		splay(x, 0); return x;
	}
	void Build(int &x, int l, int r, int pre) {
		if(l > r) return ;
		int mid = (l + r) >> 1;
		NewNode(x, s[mid], pre);
		Build(ch[x][0], l, mid-1, x);
		Build(ch[x][1], mid+1, r, x);
		up(x);
	}
} sp;
void init() {
	base[0] = 1;
	for(int i = 1; i <= MAXN; i++) base[i] = 1LL * base[i-1] * 28 % MOD;
}
int Find_RK(int x, int len) {
	int y = x + len - 1, a, b;
	a = sp.Find_kth(x-1);
	b = sp.Find_kth(y+1);
	sp.splay(a, 0); sp.splay(b, a);
	return sp.RK[sp.ch[b][0]];
}
bool check(int x, int y, int len) {
	int a = Find_RK(x, len);
	int b = Find_RK(y, len);
	return a == b;
}
void R(int x, int key) {
	int pos = sp.Find_kth(x);
	sp.splay(pos, 0); sp.val[pos] = key; sp.up(pos);
}
void I(int x, int key) {
	int a = sp.Find_kth(x), b = sp.Find_kth(x+1), cur;
	sp.splay(a, 0); sp.splay(b, a);
	sp.NewNode(cur, key, b); sp.ch[b][0] = cur;
	sp.up(cur); sp.up(b); sp.up(a);
	n++;
}
void Q(int x, int y) {
	int L = 1, R = min(n-x, n-y), ans = 0;
	while(L <= R) {
		int mid = (L + R + 1) >> 1;
		if(check(x, y, mid)) L = mid+1, ans = mid;
		else R = mid-1;
	}
	PF("%d\n", ans);
}
int main() {
	init();
	SF("%s", s+1);
	n = strlen(s+1);
	for(int i = 1; i <= n; i++) s[i] -= 'a';
	s[0] = s[n+1] = 27;
	sp.Build(sp.root, 0, n+1, 0);
	n += 2;
	SF("%d", &m);
	for(int i = 1; i <= m; i++) {
		int x, y;
		SF(" %c", &c);
		switch(c) {
			case 'R' :
				SF("%d", &x);
				if(x > n-2) continue ; x++;
				SF(" %c", &c);
				R(x, c-'a');
				break;
			case 'I' :
				SF("%d", &x); x++;
				SF(" %c", &c);
				I(x, c-'a');
				break;
			case 'Q' :
				SF("%d%d", &x, &y); x++; y++;
				Q(x, y);
		}
	}
}