Submission #336796

#	Time	Username	Problem	Language	Result	Execution time	Memory
336796		super_j6	Abduction 2 (JOI17_abduction2)	C++14	100 / 100	1032 ms	101608 KiB

abduction2

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <array>
#include <unordered_map>
using namespace std;
#define endl '\n'
#define ll long long
#define pi pair<int, int>
#define f first
#define s second
 
typedef array<int, 2> T;
T operator+(T x, T y){ return {x[0] + y[0], x[1] + y[1]};}
T operator*(int x, T y){ return {x * y[0], x * y[1]};}
 
const int mxn = 50000, m = __lg(mxn - 1) + 1, k = 4;
const T d[k] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
int n[2], q;
int a[2][m][mxn];
unordered_map<int, ll> dp[k][mxn];
 
ll sol(T x, int y){
	ll ret = 0, s = 0;
	for(int i = m - 1; ~i; i--){
		T p = x + (1 << i) * d[y];
		if(p[y & 1] >= -1 && p[y & 1] <= n[y & 1] && a[!(y & 1)][0][p[!(y & 1)]] > 
		a[y & 1][i][y & 2 ? p[y & 1] + 1 : x[y & 1]]){
			x = p, s += (1 << i);
		}
	}
	if(!~x[y & 1] || x[y & 1] == n[y & 1]) return s;
	if(dp[y][x[0]].count(x[1])) return dp[y][x[0]][x[1]] + s;
	for(int i = 0; i < 2; i++){
		int z = (y + 2 * i + 1) % k;
		ret = max(ret, sol(x + d[z], z));
	}
	dp[y][x[0]][x[1]] = ++ret;
	return ret + s;
}
 
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n[0] >> n[1] >> q;
	
	for(int i = 0; i < 2; i++){
		for(int j = 0; j < n[i]; j++){
			cin >> a[i][0][j];
		}
		for(int j = 1; j < m; j++)
		for(int l = 0; l + (1 << j) <= n[i]; l++){
			a[i][j][l] = max(a[i][j - 1][l], a[i][j - 1][l + (1 << (j - 1))]);
		}
	}
	
	while(q--){
		T x;
		for(int i = 0; i < 2; i++) cin >> x[i], x[i]--;
		ll ret = 0;
		for(int i = 0; i < k; i++) ret = max(ret, sol(x + d[i], i));
		cout << ret << endl;
	}
 
	return 0;
}

• Subtask 1: 13 / 13
• Subtask 2: 10 / 10
• Subtask 3: 17 / 17
• Subtask 4: 4 / 4
• Subtask 5: 56 / 56