제출 #311314

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
311314		eggag32	보물 찾기 (CEOI13_treasure2)	C++17	0 / 100	2 ms	512 KiB

treasure

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include "treasure.h"
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define debug(x) cerr << #x << ": " << x << endl
#define debug2(x, y) debug(x), debug(y)
#define repn(i, a, b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(i, a) for(int i = 0; i < (int)(a); i++)
#define all(v) v.begin(), v.end() 
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define sq(x) ((x) * (x))
const int mxN = 105;

template<class T> T gcd(T a, T b){ return ((b == 0) ? a : gcd(b, a % b)); }

int n, tot;
int l[mxN], r[mxN], u[mxN], d[mxN]; //all of these are inclusive
int LU[mxN][mxN], LD[mxN][mxN], RU[mxN][mxN], RD[mxN][mxN];

int qry(int r1, int c1, int r2, int c2){
	int ret = countTreasure(r1, c1, r2, c2);
	return ret;
}

bool check(int i, int j){
	int cur = 0;
	if(i > 1) cur += u[i - 1];
	if(i < n) cur += d[i + 1];
	if(j > 1) cur += l[j - 1];
	if(j < n) cur += r[j + 1];
	//now we do the intersections
	if(i > 1 && j > 1) cur -= LU[i - 1][j - 1];
	if(i < n && j > 1) cur -= LD[i + 1][j - 1];
	if(i > 1 && j < n) cur -= RU[i - 1][j + 1];
	if(i < n && j < n) cur -= RD[i + 1][j + 1];
	//assert(cur == tot || cur == (tot - 1));
	return cur != tot;
}

void solve1(int i, int j){
	int q = qry(1, j, i, n);
	if(~RU[i][j]) assert(q == RU[i][j]);
	RU[i][j] = q;
	if(j > 1) LU[i][j - 1] = u[i] - RU[i][j];
	if(i < n){
		RD[i + 1][j] = r[j] - RU[i][j];
		if(j > 1) LD[i + 1][j - 1] = l[j - 1] - LU[i][j - 1];
	}
}

void solve2(int i, int j){
	int q = qry(1, 1, i, j);
	if(~LU[i][j]) assert(q == LU[i][j]);
	LU[i][j] = q;
	if(i < n) LD[i + 1][j] = l[j] - LU[i][j];
	if(j < n){
		RU[i][j + 1] = u[i] - LU[i][j];
		if(i < n) RD[i + 1][j + 1] = r[j + 1] - RU[i][j + 1];
	}
}

void solve3(int i, int j){
	int q = qry(i, j, n, n);
	if(~RD[i][j]) assert(q == RD[i][j]);
	RD[i][j] = q;
	if(i > 1) RU[i - 1][j] = r[j] - RD[i][j];
	if(j > 1){
		LD[i][j - 1] = d[i] - RD[i][j];
		if(i > 1) LU[i - 1][j - 1] = l[j - 1] - LD[i][j - 1];
	}
}

void solve4(int i, int j){
	int q = qry(i, 1, n, j);
	if(~LD[i][j]) assert(LD[i][j] == q);
	LD[i][j] = q;
	if(i > 1) LU[i - 1][j] = l[j] - LD[i][j];
	if(j < n){
		RD[i][j + 1] = d[i] - LD[i][j];
		if(i > 1) RU[i - 1][j + 1] = r[j + 1] - RD[i][j + 1];
	}
}

void findTreasure(int N){
	n = N;
	memset(LD, -1, sizeof(LD));
	memset(LU, -1, sizeof(LU));
	memset(RD, -1, sizeof(RD));
	memset(RU, -1, sizeof(RU));
	tot = qry(1, 1, n, n);
	//l and r
	l[n] = r[1] = tot;
	repn(i, 1, n){
		if(i > (n - i)) l[i] = qry(1, 1, n, i);
		else l[i] = tot - qry(1, i + 1, n, n);
	}
	repn(i, 2, n + 1) r[i] = tot - l[i - 1];
	//u and d
	u[n] = d[1] = tot;
	repn(i, 1, n){
		if(i > (n - i)) u[i] = qry(1, 1, i, n);
		else u[i] = tot - qry(i + 1, 1, n, n);
	}
	repn(i, 2, n + 1) d[i] = tot - u[i - 1];
	//corners
	repn(i, 1, n + 1){
		repn(j, 1, n + 1){
			if(i < (n - i)){
				if(j < (n - j)) solve3(i, j);
				else solve4(i, j);
			}
			else{
				if(j < (n - j)) solve1(i, j);
				else solve2(i, j);
			}
		}
	}	
	repn(i, 1, n + 1){
		repn(j, 1, n + 1){
			int f = 1;
			if(~LD[i][j]) f = 1;
			if(~RD[i][j]) f = 1;
			if(~LU[i][j]) f = 1;
			if(~RU[i][j]) f = 1;
			if(f){
				solve1(i, j);
				solve2(i, j);
				solve3(i, j);
				solve4(i, j);
			}
		}
	}
	repn(i, 1, n + 1){
		repn(j, 1, n + 1){
			if(check(i, j)) Report(i, j);
		}
	}
}
/*
Things to look out for:
	- Integer overflows
	- Array bounds
	- Special cases
Be careful!
*/

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