Submission #17629

#	Time	Username	Problem	Language	Result	Execution time	Memory
17629		gs14004	Adriatic (CEOI13_adriatic)	C++14	60 / 100	2000 ms	28224 KiB

adriatic

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <limits.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <string>
#include <functional>
#include <vector>
#include <numeric>
#include <deque>
#include <bitset>
#include <iostream>
using namespace std;
typedef long long lint;
typedef long double llf;
typedef pair<int, int> pi;
const int S = 2500;

int n, arr[2505][2505];
pi a[250005];

int getsum(int sx, int ex, int sy, int ey){
	return arr[ex][ey] - arr[sx-1][ey] - arr[ex][sy-1] + arr[sx-1][sy-1];
}

int prec1[2505], prec2[2505], prec3[2505], prec4[2505];

void contract1(int &x, int &y){
	int rx = max(x, prec2[y+1]), ry = min(y, prec1[x-1]);
	x = rx, y = ry;
}

void contract2(int &x, int &y){
	int rx = min(x, prec4[y-1]), ry = max(y, prec3[x+1]);
	x = rx, y = ry;
}

int main(){
	scanf("%d",&n);
	for(int i=0; i<n; i++){
		scanf("%d %d",&a[i].first, &a[i].second);
		arr[a[i].first][a[i].second]++;
	}
	memset(prec1, 0x3f, sizeof(prec1));
	memset(prec4, 0x3f, sizeof(prec4));
	for(int i=0; i<n; i++){
		prec1[a[i].first] = min(prec1[a[i].first], a[i].second);
		prec2[a[i].second] = max(prec2[a[i].second], a[i].first);
		prec3[a[i].first] = max(prec3[a[i].first], a[i].second);
		prec4[a[i].second] = min(prec4[a[i].second], a[i].first);
	}
	for(int i=1; i<=S; i++){
		prec1[i] = min(prec1[i-1], prec1[i]);
		prec2[S+1-i] = max(prec2[S+1-i], prec2[S+2-i]);
		prec3[S+1-i] = max(prec3[S+1-i], prec3[S+2-i]);
		prec4[i] = min(prec4[i-1], prec4[i]);
	}
	for(int i=1; i<=S; i++){
		for(int j=1; j<=S; j++){
			arr[i][j] += arr[i-1][j] + arr[i][j-1] - arr[i-1][j-1];
		}
	}
	for(int i=0; i<n; i++){
		int x1, y1, x2, y2;
		tie(x1, y1) = a[i], tie(x2, y2) = a[i];
		lint ret = n - 3;
		for(int t=0; t<n; t++){
			ret += getsum(x1, S, 1, y1) + getsum(1, x2, y2, S);
			contract1(x1, y1);
			contract2(x2, y2);
		}
		printf("%lld\n",ret);
	}
}

• Subtask 1: 25 / 25
• Subtask 2: 25 / 25
• Subtask 3: 10 / 10
• Subtask 4: 0 / 20
• Subtask 5: 0 / 20