【基础算法：排序】总结

前言

通过排序，我们可以将数据有序化，处理数据更方便

对于各种排序算法只需了解原理即可，重点学习C++中的sort()函数
排序算法可视化
C++ sort()排序详解

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模板题 难度：橙

题意

思路

sort秒了

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;

int n, a[N];
int main() {
	cin >> n;
	for (int i = 0; i < n; ++i) cin >> a[i];

	sort(a, a + n);

	for (int i = 0; i < n; ++i) cout << a[i] << ' ';
	return 0;
}

下面是几道简单例题

[NOIP2006 普及组] 明明的随机数 难度：橙

去重排序

题意

思路

排序+去重，排序可以用sort，去重更简单了

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 1e2 + 5;

int n, cnt;
int a[N];
int main() {
	cin >> n;
	for (int i = 0; i < n; ++i) cin >> a[i];

	sort(a, a + n);
	for (int i = 0; i < n; i++)	if (a[i] != a[i + 1]) cnt++;

	cout << cnt << endl;
	for (int i = 0; i < n; ++i) {
		if (a[i] == a[i + 1]) continue;
		cout << a[i] << ' ';
	}
	return 0;
}

攀爬者 难度：橙

利用简单自定义排序解决问题

题意

思路

先排序，再算距离

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 5e4 + 5;

struct node {
	int x, y, z;
} a[N];
bool cmp(node x, node y) {
	return x.z < y.z;
}
double dis(node a, node b) {
	return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2) + pow(a.z - b.z, 2));
}
int n;
double ans;
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> a[i].x >> a[i].y >> a[i].z;

	sort(a + 1, a + n + 1, cmp);
	for (int i = 2; i <= n; i++) ans += dis(a[i], a[i - 1]);

	printf("%.3lf", ans);
	return 0;
}

生日 难度：橙

稍微复杂点的自定义排序

题意

思路

比年龄的话，就直接按时间排序呗

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 1e2 + 5;

struct node {
	string name;
	int year;
	int mouth;
	int day;
	int id;
} s[N];
bool cmp(node s1, node s2) {
	if (s1.year != s2.year) return s1.year < s2.year;
	else if (s1.mouth != s2.mouth) return s1.mouth < s2.mouth;
	else if (s1.day != s2.day) return s1.day < s2.day;
	else return s1.id > s2.id;
}
int n;
int main() {
	cin >> n;
	for (int i = 1; i <= n; ++i) {
		cin >> s[i].name >> s[i].year >> s[i].mouth >> s[i].day;
		s[i].id = i;
	}

	sort(s + 1, s + n + 1, cmp);

	for (int i = 1; i <= n; ++i) cout << s[i].name << endl;
	return 0;
}

帮贡排序 难度：黄

比较复杂的排序

题意

自己看题

思路

按题目要求排两遍，可以把职位映射成数字，方便排序

代码

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e3 + 5;

int n;
struct node {
	string na, zw, now;
	ll bg, le;
	int id;
} a[N];
int res(string a) {
	if (a == "BangZhu") return 0;
	if (a == "FuBangZhu") return 1;
	if (a == "HuFa") return 2;
	if (a == "ZhangLao") return 3;
	if (a == "TangZhu") return 4;
	if (a == "JingYing") return 5;
	if (a == "BangZhong") return 6;
}
string newzw(int i) {
	if (i == 1) return "BangZhu";
	if (i <= 3) return "FuBangZhu";
	if (i <= 5) return "HuFa";
	if (i <= 9) return "ZhangLao";
	if (i <= 16) return "TangZhu";
	if (i <= 41) return "JingYing";
	return "BangZhong";
}
int cmp1(node x, node y) {
	if (x.bg == y.bg) return x.id < y.id;
	else return x.bg > y.bg;
}
int cmp2(node x, node y) {
	if (x.now != y.now) return res(x.now) < res(y.now);
	else if (x.le != y.le) return x.le > y.le;
	else return x.id < y.id;
}

int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i].na >> a[i].zw >> a[i].bg >> a[i].le;
		a[i].id = i;
	}

	sort(a + 4, a + 1 + n, cmp1);
	for (int i = 1; i <= n; i++) a[i].now = newzw(i);
	sort(a + 1, a + 1 + n, cmp2);

	for (int i = 1; i <= n; i++) cout << a[i].na << " " << a[i].now << " " << a[i].le << endl;

	return 0;
}