Submission #104469

#	Time	Username	Problem	Language	Result	Execution time	Memory
104469		eriksuenderhauf	Robots (IOI13_robots)	C++11	100 / 100	1825 ms	24960 KiB

robots

#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#include "robots.h"
//#include "grader.h"
#define enl printf("\n")
#define case(t) printf("Case #%d: ", (t))
#define ni(n) scanf("%d", &(n))
#define nl(n) scanf("%I64d", &(n))
#define nai(a, n) for (int i = 0; i < (n); i++) ni(a[i])
#define nal(a, n) for (int i = 0; i < (n); i++) nl(a[i])
#define pri(n) printf("%d\n", (n))
#define prl(n) printf("%I64d\n", (n))
#define pii pair<int, int>
#define pll pair<long long, long long>
#define vii vector<pii>
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long int ll;
const double pi = acos(-1);
const int MOD = 1e9 + 9;
const int INF = 1e9 + 7;
const int MAXN = 1e6 + 5;
const double eps = 1e-9;
int x[MAXN], y[MAXN];
pii arr[MAXN];
int a, b, t;

bool ok(int cnt) {
	priority_queue<int> act;
	int l = 0;
	for (int i = 0; i < a; i++) {
		while (l < t && x[i] > arr[l].fi)
			act.push(arr[l++].se);
		for (int j = 0; j < cnt; j++) {
			if (act.empty()) break;
			act.pop();
		}
	}
	while (l < t) act.push(arr[l++].se);
	for (int i = b-1; i >= 0; i--) {
		for (int j = 0; j < cnt; j++) {
			if (act.empty()) return true;
			if (act.top() >= y[i]) return false;
			act.pop();
		}
	}
	if (act.empty())
		return true;
	return false;
}

//w<x;s<y
int putaway(int A, int B, int T, int X[], int Y[], int w[], int s[]) {
	a = A, b = B, t = T;
	int mxX = 0, mxY = 0;
	for (int i = 0; i < a; i++) {
		x[i] = X[i];
		mxX = max(mxX, X[i]);
	}
	sort(x, x + a);
	for (int i = 0; i < b; i++) {
		y[i] = Y[i];
		mxY = max(mxY, Y[i]);
	}
	sort(y, y + b);
	for (int i = 0; i < t; i++) {
		arr[i] = {w[i], s[i]};
		if (w[i] >= mxX && s[i] >= mxY)
			return -1;
	}
	sort(arr, arr + t);
	int lo = 1, hi = t, ret = t;
	while (lo <= hi) {
		int mi = (lo + hi) / 2;
		if (ok(mi))
			hi = mi - 1, ret = mi;
		else
			lo = mi + 1;
	}
	return ret;
}

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 25 / 25
• Subtask 4: 37 / 37
• Subtask 5: 10 / 10