分治 堆

//一直前序和中序求后序
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1e6 + 100;
 
int pr[maxn], in[maxn], n;          /*pr储存前序遍历，in储存中序遍历*/ 
int Left[maxn], Right[maxn], root;  //二叉树相关
bool ok;
 
//根据前序和中序建树
int build(int L1, int R1, int L2, int R2) 
{
    if(L1 > R1) 
        return 0;
    int rt = pr[L1];
    int p = 1;
    while(in[p] != rt) 
        ++p;                                          //在中序中找到根结点
    Left[rt] = build(L1+1, L1 + p-L2, L2, p-1);      //递归建立左子树
    Right[rt] = build(L1+ p-L2 + 1, R1, p+1, R2);    //递归建立右子树
    return rt;                                       //返回根结点
}
 
//后序遍历
void poOrder(int rt) 
{
    if(Left[rt] != 0) 
        poOrder(Left[rt]);
    if(Right[rt] != 0) 
        poOrder(Right[rt]);
    if(ok) 
        cout << " "; 
    if(!ok) ok = 1;
        cout << rt;
}
 
int main() 
{
    while(scanf("%d", &n) == 1) 
    {
        //初始化+输入
        memset(Left, 0, sizeof(Left)); 
        memset(Right, 0, sizeof(Right));
        for(int i = 1; i <= n; ++i) 
            scanf("%d", &pr[i]);
        for(int i = 1; i <= n; ++i) 
            scanf("%d", &in[i]);
 
        root = build(1,n,1,n);  //建树
        ok = 0;
        poOrder(root);          //后序遍历输出结果
        cout << endl;
    }
    return 0;
}

//根据树的后序和中序，求层次遍历。
#include<bits/stdc++.h>
using namespace std;
 
const int maxn = 30;
int a[maxn + 10], b[maxn + 10];
map<int, int>L, R; //L[i]表示i的左子节点,R[i]表示i的右子节点
 
//当前考虑的是后序遍历[la, ra]区间和对应的中序遍历[lb,rb]区间
int build(int la, int ra, int lb, int rb)
{
    if (la > ra)   //越界判断
    {
        return 0;
    }
    int root = a[ra]; //后序遍历右一肯定是根
    int i;
    for (i = lb; i <= rb && b[i] != root; i++) {} //在中序遍历中找到这个节点下标i
    if (i <= rb)   //递归分治
    {
        R[root] = build(ra - rb + i, ra - 1, i + 1, rb);
        L[root] = build(la, ra - rb + i - 1, lb, i - 1);
    }
    return root;
}
 
void bfs(int root)
{
    queue<int> Q;
    Q.push(root);
 
    int cnt = 0;
    while (!Q.empty())
    {
        int tn = Q.front();
        Q.pop();
        printf(cnt++ == 0 ? "%d" : " %d", tn);
        if (L[tn])
        {
            Q.push(L[tn]);
        }
        if (R[tn])
        {
            Q.push(R[tn]);
        }
    }
    puts("");
}
 
int main()
{
    int N;
    scanf("%d", &N);
    for (int i = 1; i <= N; i++)
    {
        scanf("%d", &a[i]);
    }
    for (int i = 1; i <= N; i++)
    {
        scanf("%d", &b[i]);
    }
    int root = build(1, N, 1, N);
    bfs(root);
    return 0;
}
 

//根据树的中序和按层遍历，求先序遍历。
#include<iostream>
#include<cstring>
using namespace std;
string s1,s2;
void calc(int l1,int r1,int l2,int r2)
{
    int i,j;
    for(i=l2;i<=r2;i++)//找层次遍历中优先输出根节点的位置 ，根肯定在儿子的前面。
    {
        int b=0;
        for(j=l1;j<=r1;j++)
        {
            if(s2[i]==s1[j]) 
            {
                cout<<s1[j];//输出根节点 
                b=1;
                break;
            }
        }
        if(b) break;
    }
    if(j>l1) calc(l1,j-1,0,r2);//遍历左子树 
    if(j<r1) calc(j+1,r1,0,r2);//遍历右子树 
}
int main()
{
    cin>>s1>>s2;
    calc(0,s1.length()-1,0,s2.length()-1);
    cout<<endl;
    return 0;
} 

//根据中序遍历和层次遍历，求后序遍历
#include<stdio.h>
#include<string.h>
int len,mark[102];                                                             //用来给中序遍历做标记
char s1[102],s2[102];                                                    //储存遍历
void tree(int l,int r){                                                          //递归函数
    if(l>r)return;                                                               //返回条件，不能等于
    int i,j,min=0x7fffff/*int范围的最大值*/,root;
    for(i=l;i<=r;i++){
        if(mark[s1[i]]<min){
            min=mark[s1[i]];root=i;                                        //在中序遍历中找到层次遍历里的最靠前的结点（即先序遍历中要求优先输出的根结点
        }         
    }   
    printf("%c",s1[root]);                                                  //输出根结点
    tree(l,root-1);                                                             //先遍历左子树
    tree(root+1,r);                                                              //再遍历右子树，顺序不能调换
}
int main(){
    scanf("%s%s",s1+1,s2+1)；
    len=strlen(s2+1);
    int i;
    for(i=1;i<=len;i++)
        mark[s2[i]]=i;                                                             //做标记，虽然s2[i]为字符，但视为ASCII码值，越靠前标记值越小
    tree(1,len);    
}

https://www.cnblogs.com/jhz033/p/5571233.html 刷围墙，一道分治的题目，可以用dp;

 

https://www.cnblogs.com/helloworld-c/p/4854463.html 堆

https://blog.csdn.net/Dream_Weave/article/details/82526775

https://blog.csdn.net/qq_38779698/article/details/79759662 这个比较简单。

https://pintia.cn/problem-sets/1104020079991836672/problems/1104020194857046027 题目链接

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;

struct heap {
int sum, num[1005];
int top() {
return num[1];
}
bool empty() {
if(sum == 0) return true;
return false;
}
void push(int a) {
num[++sum] = a;
int now = sum, temp;
while(now > 1 && num[now] > num[now / 2]) {
    temp = num[now];
    num[now] = num[now / 2];
    num[now / 2] = temp;
    now = now / 2;
}
}
void pop() {
int now = 1, temp;
num[now] = num[sum--];
while((now * 2 <= sum && num[now] < num[now * 2]) || (now * 2 + 1 <= sum && num[now] < num[now * 2 + 1])) {

    if(num[now * 2] > num[now * 2 + 1]) {
        temp = num[now];
        num[now] = num[now * 2];
        num[now * 2] = temp;
        now = now * 2;
    }
    else {
        temp = num[now];
        num[now] = num[now * 2 + 1];
        num[now * 2 + 1] = temp;
        now = now * 2 + 1;
    }
}
}
}a;

int main() {
int n;
cin >> n;
int b;
for(int i = 1; i <= n; i++) {
    cin >> b;
    a.push(b);
}
while(!a.empty()) {
    cout << a.top() << " ";
    a.pop();
}
return 0;
}