洛谷 P1102 A-B 数对

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#c++ #二分法 #双指针 #算法

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本文介绍了洛谷P1102问题的两种高效解法，包括使用二分查找法在100毫秒以上的复杂度下解决问题，以及利用双指针法在100毫秒以内实现解决方案。这两种算法在解决数对问题上展现了不同的思路和效率。

P1102 A-B 数对

二分查找解法(100+ms)

#include <cstdio>
#include <algorithm>
#define MAXN 200020
using namespace std;
int a[MAXN];
inline int read() {
	int x = 0, sign = 0;
	char c = getchar();
	while (c < '0' || c > '9') {
		if ('-' == c) {
			sign = 1;
		}
		c = getchar();
	}
	while ('0' <= c && c <= '9') {
		x = (x << 3) + (x << 1) + (c ^ 0x30);
		c = getchar();
	}
	return sign ? (~x + 1) : x;
}
int main() {
	int N = read();
	int C = read();
	long ans = 0;
	for (register int i = 0; i < N; ++i) {
		a[i] = read();
	}
	sort(a, a + N);
	for (register int i = 0; i < N; ++i) {
		ans += upper_bound(a, a + N, a[i] + C) - lower_bound(a, a + N, a[i] + C) ;
	}
	printf("%ld", ans);
	return 0;
}

双指针解法(100-ms)

#include <cstdio>
#include <algorithm>
#define MAXN 200020
using namespace std;
int a[MAXN];
inline int read() {
	int x = 0, sign = 0;
	char c = getchar();
	while (c < '0' || c > '9') {
		if ('-' == c) {
			sign = 1;
		}
		c = getchar();
	}
	while ('0' <= c && c <= '9') {
		x = (x << 3) + (x << 1) + (c ^ 0x30);
		c = getchar();
	}
	return sign ? (~x + 1) : x;
}
int main() {
	int N = read();
	int C = read();
	long ans = 0;
	for (register int i = 0; i < N; ++i) {
		a[i] = read();
	}
	sort(a, a + N);
	register int lo = 0;
	register int hi = 0;
	for (register int i = 0; i < N; ++i) {
		while (lo < N && a[lo] - a[i] < C) {
			++lo;
		}
		while (hi < N && a[hi] - a[i] <= C) {
			++hi;
		}
		if ((C == a[hi - 1] - a[i]) && (C == a[lo] - a[i])) {
			ans += hi - lo;
		}
	}
	printf("%ld", ans);
	return 0;
}