Codeforces Round #763 (Div. 2) B. Game on Ranges

Problem - 1623B - Codeforces

这个题就是给出一些区间，然后每次就从该区间内取出一个数，最后能够把1...n所有数都能取一遍，求每个区间取哪个数，因为数据量比较小，可以直接暴搜

AC代码：

版本1

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define lowbit(x) ((x) & -(x))
#define IOS1 ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define IOS2 ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
typedef vector vi;
typedef vector vll;
typedef vector vc;
template T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template T lcm(T a, T b) { return a / gcd(a, b) * b; }
template
T power(T a, int b) {
	T res = 1;
	for (; b; b >>= 1, a = a * a) {
		if (b & 1) {
			res = res * a;
		}
	}
	return res;
}

const int INF = 0x3f3f3f3f;
const int mod = 1000000007;

void solve() {
	int n;
	cin >> n;
	vector l(n), r(n);
	for (int i = 0; i < n; i++) {
		cin >> l[i] >> r[i];
		l[i]--;
	}
	for (int i = 0; i < n; i++) {
		int x = l[i];
		for (int j = 0; j < n; j++) {
			if (l[i] == l[j] && r[j] < r[i]) {
				x = max(x, r[j]);
			}
		}
		cout << l[i] + 1 << " " << r[i] << " " << x + 1 << endl;
	}
	
}

int main() {
	IOS1;
	//IOS2;
	int __t = 1;
	cin >> __t;
	for (int _t = 1; _t <= __t; _t++) {
		solve();
	}
	return 0;
}

版本2

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define lowbit(x) ((x) & -(x))
#define IOS1 ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define IOS2 ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
typedef vector vi;
typedef vector vll;
typedef vector vc;
template T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template T lcm(T a, T b) { return a / gcd(a, b) * b; }
template
T power(T a, int b) {
	T res = 1;
	for (; b; b >>= 1, a = a * a) {
		if (b & 1) {
			res = res * a;
		}
	}
	return res;
}

const int INF = 0x3f3f3f3f;
const int mod = 1000000007;

void solve() {
	int n;
	cin >> n;
	vector l(n), r(n);
	for (int j = 0; j < n; j++) {
		cin >> l[j] >> r[j];
	}
	set> st;
	for (int j = 0; j < n; j++) {
		st.insert(make_pair(l[j], r[j]));
	}
	for (int j = 0; j < n; j++) {
		for (int k = l[j]; k <= r[j]; k++) {
			bool ok = true;
			if (k != l[j]) {
				if (st.count(make_pair(l[j], k - 1)) == 0) {
					ok = false;
				}
			}
			if (k != r[j]) {
				if (st.count(make_pair(k + 1, r[j])) == 0) {
					ok = false;
				}
			}
			if (ok) {
				cout << l[j] << ' ' << r[j] << ' ' << k << endl;
			}
		}
	}
}

int main() {
	IOS1;
	//IOS2;
	int __t = 1;
	cin >> __t;
	for (int _t = 1; _t <= __t; _t++) {
		solve();
	}
	return 0;
}

Codeforces Round #763 (Div. 2) B. Game on Ranges

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