HDU 4125 Moles(nlogn建立二叉搜索树、kmp)

题意:

$N\le 6\times 10^5,给定1\sim N的序列,按照这个顺序建立一颗二叉搜索树$
$奇数是1,偶数是0,先序遍历这颗二叉搜索树生成1个01的欧拉序列$
$查找T串可重叠的出现了几次,|T|\le 7000$

分析:

$不学是要还的。。$
$这个数据范围只能nlogn建立BST了,set动态维护一下插入的节点$
$每次查找比当前x大的的最小点r,以及比x小的最大点l$
$分别尝试r的左儿子,以及l的右儿子能不能插入就可以了$
$然后先序遍历一遍,C++开栈一发,跑个kmp就ok了$
$时间复杂度O(nlogn)$

代码:

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//
// Created by TaoSama on 2016-05-03
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 6e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, m;
int rt, ls[N], rs[N];
char s[N << 1], t[7010];
void dfs(int rt) {
s[n++] = (rt & 1) + '0';
if(ls[rt]) {
dfs(ls[rt]);
s[n++] = (rt & 1) + '0';
}
if(rs[rt]) {
dfs(rs[rt]);
s[n++] = (rt & 1) + '0';
}
}
int kmp() {
m = strlen(t);
vector<int> nxt(m + 1);
nxt[0] = -1;
for(int i = 0, j = -1; i < m;) {
if(j == -1 || t[i] == t[j]) nxt[++i] = ++j;
else j = nxt[j];
}
int ret = 0;
for(int i = 0, j = 0; i < n;) {
if(j == -1 || s[i] == t[j]) ++i, ++j;
else j = nxt[j];
if(j == m) {
++ret;
j = nxt[j];
}
}
return ret;
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int T; scanf("%d", &T);
while(T--) {
scanf("%d", &n);
memset(ls, 0, sizeof ls);
memset(rs, 0, sizeof rs);
set<int> st;
for(int i = 1; i <= n; ++i) {
int x; scanf("%d", &x);
if(!st.size()) rt = x;
else {
auto it = st.lower_bound(x);
if(it == st.end()) rs[*--it] = x;
else {
if(!ls[*it]) ls[*it] = x;
else rs[*--it] = x;
}
}
st.insert(x);
}
scanf("%s", t);
n = 0;
dfs(rt);
s[n] = 0;
static int kase = 0;
printf("Case #%d: %d\n", ++kase, kmp());
}
return 0;
}


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