제출 #56590

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
56590		Xellos	Snake Escaping (JOI18_snake_escaping)	C++14	12 / 100	645 ms	66560 KiB

snake_escaping

#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
// iostream is too mainstream
#include <cstdio>
// bitch please
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <cmath>
#include <iomanip>
#include <time.h>
#define dibs reserve
#define OVER9000 1234567890123456789LL
#define ALL_THE(CAKE,LIE) for(auto LIE =CAKE.begin(); LIE != CAKE.end(); LIE++)
#define tisic 47
#define soclose 1e-8
#define chocolate win
// so much chocolate
#define patkan 9
#define ff first
#define ss second
#define abs(x) ((x < 0)?-(x):x)
#define uint unsigned int
#define dbl long double
#define pi 3.14159265358979323846
using namespace std;
// mylittledoge
 
typedef long long cat;
 
#ifdef DONLINE_JUDGE
	// palindromic tree is better than splay tree!
	#define lld I64d
#endif
 
int cost_arr[2][1<<19];
int cost[1<<20];
int ans[1000000];
cat query[1000000];

constexpr int pw3(int e) {
	if(e <= 0) return 1;
	return 3 * pw3(e-1);
}

constexpr int C = 10;
int v[pw3(C)];
int ids[pw3(C)][2];

constexpr void compute_ids() {
	for(int i = 0; i < pw3(C); i++) {
		int x = i, y = 0;
		for(int j = 0; j <= C; j++) {
			if(j == C) {
				ids[i][0] = ids[i][1] = y;
				break;
			}
			if(x%3 == 2) {
				ids[i][0] = i-pw3(j), ids[i][1] = i-2*pw3(j);
				break;
			}
			y += (1<<j) * (x%3);
			x /= 3;
		}
	}
}

void solve(int d, int B, int cl, int cr, int ql, int qr) {
	if(ql == qr) return;
	if(d == B-C) {
		for(int i = 0; i < pw3(C); i++) {
			if(ids[i][0] == ids[i][1]) v[i] = cost[cl+ids[i][0]];
			else v[i] = v[ids[i][0]] + v[ids[i][1]];
		}
		for(int i = ql; i < qr; i++) {
			int w = 0;
			for(int j = C-1; j >= 0; j--) {
				if((query[i]>>(20+d+j))&1) w = 3 * w + ((query[i]>>(d+j))&1);
				else w = 3 * w + 2;
			}
			ans[(query[i]>>40)&((1LL<<20)-1)] += ((query[i]>>60) ? -v[w] : v[w]);
		}
		return;
	}
	if(d == B) {
		for(int i = ql; i < qr; i++) 
			ans[(query[i]>>40)&((1LL<<20)-1)] += ((query[i]>>60) ? -cost[cl] : cost[cl]);
		return;
	}
	int cnt[3] = {};
	for(int i = ql; i < qr; i++) {
		if((query[i]>>(20+d))&1) cnt[(query[i]>>(d))&1]++;
		else cnt[2]++;
	}
	if(cnt[2] <= cnt[0] && cnt[2] <= cnt[1]) {
		for(int i = 0; i < (1<<(B-1-d)); i++) {
			cost_arr[0][i] = cost[cl+2*i];
			cost_arr[1][i] = cost[cl+2*i+1];
		}
		for(int i = 0; i < (1<<(B-1-d)); i++) cost[cl+i] = cost_arr[0][i];
		for(int i = 0; i < (1<<(B-1-d)); i++) cost[cl+(1<<(B-1-d))+i] = cost_arr[1][i];
		int a = ql, b = ql+cnt[0]+cnt[2];
		cat bits = (1LL<<(20+d)) + (1LL<<(d));
		for(int i = ql; i < ql+cnt[0]+cnt[2]; i++) {
			if((query[i]>>(20+d))&1) {
				if((query[i]>>(d))&1) {
					for(; b < qr && (query[b]&bits) == bits; b++) {}
					if(b == qr) break;
					swap(query[i], query[b]);
					b++;
				}
				else {
					while(a < i && ((query[a]>>(20+d))&1) == 1 && ((query[a]>>(d))&1) == 0) a++;
					if(a == i) continue;
					swap(query[i], query[a]);
					a++;
				}
			}
		}
		solve(d+1, B, cl, cl+(1<<(B-1-d)), ql, ql+cnt[0]+cnt[2]);
		a = ql;
		for(int i = ql+cnt[0]; i < ql+cnt[0]+cnt[2]; i++) if(((query[i]>>(20+d))&1) == 1) {
			for(; a < ql+cnt[0] && ((query[a]>>(20+d))&1) == 1; a++) {}
			if(a == ql+cnt[0]) break;
			swap(query[i], query[a]);
			a++;
		}
		solve(d+1, B, cl+(1<<(B-1-d)), cr, qr-cnt[2]-cnt[1], qr);
		return;
	}
	int x = (int) (cnt[1] < cnt[0]);
	for(int i = 0; i < (1<<(B-1-d)); i++) {
		cost_arr[0][i] = cost[cl+2*i+1-x];
		cost_arr[1][i] = cost[cl+2*i] + cost[cl+2*i+1];
	}
	for(int i = 0; i < (1<<(B-1-d)); i++) cost[cl+i] = cost_arr[0][i];
	for(int i = 0; i < (1<<(B-1-d)); i++) cost[cl+(1<<(B-1-d))+i] = cost_arr[1][i];
	int a = ql, b = ql+cnt[1-x]+cnt[x];
	for(int i = ql; i < ql+cnt[1-x]+cnt[x]; i++) {
		if((query[i]>>(20+d))&1) {
			if(((query[i]>>(d))&1) != x) {
				while(a < i && ((query[a]>>(20+d))&1) == 1 && ((query[a]>>(d))&1) != x) a++;
				if(a == i) continue;
				swap(query[i], query[a]);
				a++;
			}
		}
		else {
			while(b < qr && ((query[b]>>(20+d))&1) == 0) b++;
			swap(query[i], query[b]);
			b++;
			i--;
		}
	}
	for(int i = ql+cnt[1-x]; i < ql+cnt[1-x]+cnt[x]; i++) query[i] ^= 1LL<<60;
	solve(d+1, B, cl, cl+(1<<(B-1-d)), ql, ql+cnt[1-x]+cnt[x]);
	a = ql+cnt[1-x];
	cat bits = (1LL<<(20+d)) + (1LL<<(d));
	cat expect = (1LL<<(20+d)) + x * (1LL<<(d));
	for(int i = ql; i < ql+cnt[1-x]; i++) if((query[i]&bits) == expect) {
		while(a < ql+cnt[1-x]+cnt[x] && (query[a]&bits) == expect) a++;
		if(a == ql+cnt[1-x]+cnt[x]) break;
		swap(query[i], query[a]);
		a++;
	}
	for(int i = ql+cnt[1-x]; i < ql+cnt[1-x]+cnt[x]; i++) query[i] ^= 1LL<<60;
	solve(d+1, B, cl+(1<<(B-1-d)), cr, qr-cnt[2]-cnt[x], qr);
}

int main() {
	compute_ids();
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(10);
	int B, Q;
	char buf[(1<<20) + 10];
	scanf("%d %d %s", &B, &Q, buf);
	for(int i = 0; i < (1<<B); i++) {
		int b = 0;
		for(int j = 0; j < B; j++) b = 2*b + ((i>>j)&1);
		cost[b] = buf[i]-'0';
	}
	// query: mask, bits, id
	char buf_small[30];
	for(int i = 0; i < Q; i++) {
		scanf("%s", buf_small);
		query[i] = (1LL*i)<<40;
		for(int j = 0; j < B; j++) {
			if(buf_small[j] == '?') continue;
			query[i] += 1LL<<(j+20);
			if(buf_small[j] == '1') query[i] += 1LL<<(j+0);
		}
	}
	memset(ans, 0, sizeof(ans));
	solve(0, B, 0, 1<<B, 0, Q);
	for(int i = 0; i < Q; i++) cout << ans[i] << "\n";
	return 0;}
 
// look at my code
// my code is amazing

컴파일 시 표준 에러 (stderr) 메시지

snake_escaping.cpp: In function 'int main()':
snake_escaping.cpp:184:7: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
  scanf("%d %d %s", &B, &Q, buf);
  ~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~
snake_escaping.cpp:193:8: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
   scanf("%s", buf_small);
   ~~~~~^~~~~~~~~~~~~~~~~

• Subtask 1: 5 / 5
• Subtask 2: 7 / 7
• Subtask 3: 0 / 10
• Subtask 4: 0 / 53
• Subtask 5: 0 / 25