제출 #534326

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
534326		ac2hu	무제 (POI11_tem)	C++14	50 / 100	205 ms	25960 KiB

Main

#include <bits/stdc++.h>
#ifdef DEBUG
#include "../templates/debug.h"
#else 
#define deb(x...)
#endif
using namespace std;
const int N = 1e6 + 10;
int L[N],R[N];
int leftmost[N];
char in[(1 << 23)];
template<typename T> struct rmq {
	int n;
	static const int b = 30; // block size
	vector<int> mask, t; // mask and sparse table
 
	int op(int x, int y) {
		return L[x] > L[y] ? x : y;
	}
	// least significant set bit
	int lsb(int x) {
		return x & -x;
	}
	// index of the most significant set bit
	int msb_index(int x) {
		return __builtin_clz(1)-__builtin_clz(x);
	}
	// answer query of v[r-size+1..r] using the masks, given size <= b
	int small(int r, int size = b) {
		// get only 'size' least significant bits of the mask
		// and then get the index of the msb of that
		int dist_from_r = msb_index(mask[r] & ((1<<size)-1));
 
		return r - dist_from_r;
	}
	rmq(int n) : n(n), mask(n), t(n) {
		int curr_mask = 0;
		for (int i = 0; i < n; i++) {
 
			// shift mask by 1, keeping only the 'b' least significant bits
			curr_mask = (curr_mask<<1) & ((1<<b)-1);
 
			while (curr_mask > 0 and op(i, i - msb_index(lsb(curr_mask))) == i) {
				// current value is smaller than the value represented by the
				// last 1 in curr_mask, so we need to turn off that bit
				curr_mask ^= lsb(curr_mask);
			}
			// append extra 1 to the mask
			curr_mask |= 1;
 
			mask[i] = curr_mask;
		}
 
		// build sparse table over the n/b blocks
		// the sparse table is linearized, so what would be at
		// table[j][i] is stored in table[(n/b)*j + i]
		for (int i = 0; i < n/b; i++) t[i] = small(b*i+b-1);
		for (int j = 1; (1<<j) <= n/b; j++) for (int i = 0; i+(1<<j) <= n/b; i++)
			t[n/b*j+i] = op(t[n/b*(j-1)+i], t[n/b*(j-1)+i+(1<<(j-1))]);
	}
	// query(l, r) returns the actual minimum of v[l..r]
	// to get the index, just change the first and last lines of the function
	T query(int l, int r) {
		// query too small
		if (r-l+1 <= b) return L[small(r, r-l+1)];
 
		// get the minimum of the endpoints
		// (there is no problem if the ranges overlap with the sparse table query)
		int ans = op(small(l+b-1), small(r));
 
		// 'x' and 'y' are the blocks we need to query over
		int x = l/b+1, y = r/b-1;
 
		if (x <= y) {
			int j = msb_index(y-x+1);
			ans = op(ans, op(t[n/b*j+x], t[n/b*j+y-(1<<j)+1]));
		}
 
		return L[ans];
	}
};
struct scanner
{
    char const* o;
    scanner(): o(in){ load(); }
    void load(){ in[ fread(in , 1, sizeof(in)- 4, stdin)] = 0; }
    long long  readInt(){
        unsigned long long  u = 0, s = 0;
        while(*o && *o <= 32)++o;
        if (*o == '-')s = ~s, ++o; else if (*o == '+')++o;
        while(*o >='0' && *o <='9') u = (u << 3) + (u << 1) + (*o++ - '0');
        return (u^s) + !!s;
    }
};
signed main() {
	iostream::sync_with_stdio(false);
	cin.tie(nullptr);cout.tie(nullptr);
	scanner sc;
	int n = sc.readInt();
	for(int i = 0;i<n;i++){
		L[i] = sc.readInt();
		R[i] = sc.readInt();
	}
	rmq<int> rq(n);
	for(int i =0;i<n;i++){
		int l = 0,r = i;
		while(l < r){
			int mid = (l + r)/2;
			if(rq.query(mid,i) <= R[i])
				r = mid;
			else
				l = mid + 1;
		}
		leftmost[i] = l;
	}
	for(int i = 1;i<n;i++)
		leftmost[i] = max(leftmost[i], leftmost[i - 1]);
	int mx= 0 ;
	for(int i = 0;i<n;i++){
		mx = max(mx, i - leftmost[i] + 1);
	}
	cout << mx << "\n";
}

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