要求
给定n个数，对这n个数进行排序
这题当然可以直接调用sort
#include<cstdio>
#include<vector>
#define ll long long
using namespace std;
ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n;
vector<int> a;
int main()
{
n=read();
for(int i=1;i<=n;i++)
{
int x=read();
a.push_back(x);
}
sort(a.begin(),a.end());
for(vector<int>::iterator i=a.begin();i!=a.end();i++)
printf("%d ",*i);
return 0;
}
用set实现排序，元素必须无重复
1 #include<cstdio>
2 #include<set>
3 #define ll long long
4 using namespace std;
5 ll read()
6 {
7 ll x=0,f=1;char ch=getchar();
8 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
9 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
10 return x*f;
11 }
12 int n;
13 set<int>st;
14 int main()
15 {
16 n=read();
17 for(int i=1;i<=n;i++)
18 {
19 int x=read();
20 st.insert(x);
21 }
22 for(set<int>::iterator i=st.begin();i!=st.end();i++)
23 printf("%d ",*i);
24 return 0;
25 }
用二叉搜索树来排序，但不能通过已经排序好的大数据点
1 #include<cstdio>
2 #define ll long long
3 using namespace std;
4 ll read()
5 {
6 ll x = 0, f = 1; char ch = getchar();
7 while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
8 while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
9 return x * f;
10 }
11 int rt, cnt; //rt为根节点标号，cnt为当前节点个数
12 int t, n, ans;
13 int v[200005], ls[200005], rs[200005];
14 int insert(int &k, int x)
15 {
16 if (!k)
17 {
18 k = ++cnt;
19 v[k] = x;
20 return k;
21 }
22 if (x < v[k]) insert(ls[k], x);
23 else insert(rs[k], x);
24 return k;
25 }
26
27 //中序遍历
28 void dfs(int x)
29 {
30 if (!x)return;
31 dfs(ls[x]);
32 printf("%d ", v[x]);
33 dfs(rs[x]);
34 }
35 int main()
36 {
37 n = read();
38 for (int i = 1; i <= n; i++)
39 {
40 int x = read();
41 insert(rt, x);
42 }
43 dfs(rt);
44 return 0;
45 }
可以打乱输入的数据实现深度期望
1 #include<cstdio>
2 #include<cstdlib>
3 #include<algorithm>
4 #define ll long long
5 using namespace std;
6
7 ll read()
8 {
9 ll x = 0, f = 1; char ch = getchar();
10 while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
11 while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
12 return x * f;
13 }
14 int rt, cnt;
15 int t, n, ans;
16 int v[200005], ls[200005], rs[200005];
17
18 int insert(int &k, int x)
19 {
20 if (!k)
21 {
22 k = ++cnt;
23 v[k] = x;
24 return k;
25 }
26 if (x < v[k])insert(ls[k], x);
27 else insert(rs[k], x);
28 return k;
29 }
30 void dfs(int x)
31 {
32 if (!x)return;
33 dfs(ls[x]);
34 printf("%d ", v[x]);
35 dfs(rs[x]);
36 }
37 int a[200005];
38 int main()
39 {
40 n = read();
41 for (int i = 1; i <= n; i++)
42 {
43 a[i] = read();
44 swap(a[i], a[rand() % i + 1]);
45 }
46 for (int i = 1; i <= n; i++)
47 insert(rt, a[i]);
48 dfs(rt);
49 return 0;
50 }
朝鲜树，当插入超过某个深度时重构整颗树
1 #include<set>
2 #include<cmath>
3 #include<stack>
4 #include<queue>
5 #include<cstdio>
6 #include<vector>
7 #include<cstring>
8 #include<cstdlib>
9 #include<iostream>
10 #include<algorithm>
11 #define mod 1000000
12 #define pi acos(-1)
13 #define inf 0x7fffffff
14 #define ll long long
15 using namespace std;
16 ll read()
17 {
18 ll x = 0, f = 1; char ch = getchar();
19 while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
20 while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
21 return x * f;
22 }
23 bool flag;
24 int rt, cnt;
25 int t, n, ans;
26 int v[200005], ls[200005], rs[200005];
27 int a[200005];
28
29 int insert(int &k, int x, int depth)
30 {
31 if (depth > 1000) flag = 1; //插入某个数时深度大于设定，将重构标志设为true
32 if (!k)
33 {
34 k = ++cnt;
35 v[k] = x;
36 return k;
37 }
38 if (x < v[k])insert(ls[k], x, depth + 1);
39 else insert(rs[k], x, depth + 1);
40 return k;
41 }
42 void dfs(int x)
43 {
44 if (!x)return;
45 dfs(ls[x]);
46 printf("%d ", v[x]);
47 dfs(rs[x]);
48 }
49
50 //简单重构，甚至没有利用前i个有序
51 void rebuild(int &k, int l, int r)
52 {
53 if (l > r)return;
54 int mid = (l + r) >> 1;
55 k = mid;
56 v[k] = a[mid];
57 rebuild(ls[k], l, mid - 1);
58 rebuild(rs[k], mid + 1, r);
59 }
60 int main()
61 {
62 n = read();
63 for (int i = 1; i <= n; i++)
64 a[i] = read();
65 for (int i = 1; i <= n; i++)
66 {
67 insert(rt, a[i], 0);
68 if (flag)
69 {
70 for (int j = 1; j <= i; j++) ls[j] = rs[j] = v[j] = 0;
71 rebuild(rt, 1, i); //对前i个重构
72 flag = 0;
73 }
74 }
75 dfs(rt);
76 return 0;
77 }
替罪羊树
通过非旋转的重构实现的二叉平衡树，是朝鲜树的高级版，详情可见https://www.cnblogs.com/lfri/p/10006414.html
参考链接：