hdu 3999 The order of a Tree

传送门：点这里。

As we know,the shape of a binary search tree is greatly related to the order of keys we insert. To be precisely:1.  insert a key k to a empty tree, then the tree become a tree with only one node;2.  insert a key k to a nonempty tree, if k is less than the root ,insert it to the left sub-tree;else insert k to the right sub-tree.We call the order of keys we insert “the order of a tree”,your task is,given a oder of a tree, find the order of a tree with the least lexicographic order that generate the same tree.Two trees are the same if and only if they have the same shape.

Input

There are multiple test cases in an input file. The first line of each testcase is an integer n(n <= 100,000),represent the number of nodes.The second line has n intergers,k1 to kn,represent the order of a tree.To make if more simple, k1 to kn is a sequence of 1 to n.

Output

One line with n intergers, which are the order of a tree that generate the same tree with the least lexicographic.

Sample Input

41 3 4 2

Sample Output

1 3 2 4

【题意】

给你一个序列，让你按照二叉搜索树的结构建树，然后让你输出一个序列，该序列满足：1，按照二叉搜索树的结构建树之后和原树一样，2，新建立的树字典序最小。

【分析】

显然输出的是先序遍历的二叉树，这样的字典序最小。然后就根据他给的树建树，然后先序遍历一下就好。

【代码】

#include<iostream>#include<cstdio>#include<cstring>#include<string.h>#include<algorithm>#include<vector>#include<cmath>#include<stdlib.h>#include<time.h>#include<stack>#include<set>#include<map>#include<queue>#include<sstream>using namespace std;#define rep0(i,l,r) for(int i = (l);i < (r);i++)#define rep1(i,l,r) for(int i = (l);i <= (r);i++)#define rep_0(i,r,l) for(int i = (r);i > (l);i--)#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)#define MS0(a) memset(a,0,sizeof(a))#define MS_1(a) memset(a,-1,sizeof(a))#define MSinf(a) memset(a,0x3f,sizeof(a))#define MSfalse(a) memset(a,false,sizeof(a))#define inf 0x3f3f3f3f#define lson i<<1,l,mid#define rson ((i<<1)|1),mid+1,r#define uint unsigned inttypedef pair<int,int> pii;typedef pair<pii,int> piii;typedef pair<long long,long long> pll;typedef pair<pll,long long> plll;#define A first#define B second#define pb push_back#define MK make_pair#define ll long long#define eps 1e-8inline void read1(int &num) {    char in;    bool IsN=false;    in=getchar();    while(in!='-'&&(in<'0'||in>'9')) in=getchar();    if(in=='-') {        IsN=true;        num=0;    } else num=in-'0';    while(in=getchar(),in>='0'&&in<='9') {        num*=10,num+=in-'0';    }    if(IsN) num=-num;}inline void read2(int &a,int &b) {    read1(a);    read1(b);}inline void read3(int &a,int &b,int &c) {    read1(a);    read1(b);    read1(c);}inline void read1(ll &num) {    char in;    bool IsN=false;    in=getchar();    while(in!='-'&&(in<'0'||in>'9')) in=getchar();    if(in=='-') {        IsN=true;        num=0;    } else num=in-'0';    while(in=getchar(),in>='0'&&in<='9') {        num*=10,num+=in-'0';    }    if(IsN) num=-num;}inline void read2(ll &a,ll &b) {    read1(a);    read1(b);}inline void read3(ll &a,ll &b,ll &c) {    read1(a);    read1(b);    read1(c);}inline void read1(double &num) {    char in;    double Dec=0.1;    bool IsN=false,IsD=false;    in=getchar();    while(in!='-'&&in!='.'&&(in<'0'||in>'9'))        in=getchar();    if(in=='-') {        IsN=true;        num=0;    } else if(in=='.') {        IsD=true;        num=0;    } else num=in-'0';    if(!IsD) {        while(in=getchar(),in>='0'&&in<='9') {            num*=10;            num+=in-'0';        }    }    if(in!='.') {        if(IsN) num=-num;        return ;    } else {        while(in=getchar(),in>='0'&&in<='9') {            num+=Dec*(in-'0');            Dec*=0.1;        }    }    if(IsN) num=-num;}inline void read2(double &a,double &b) {    read1(a);    read1(b);}inline void read3(double &a,double &b,double &c) {    read1(a);    read1(b);    read1(c);}///-------------------------------------------------------------------------const int maxm = 1e5+10;int l[maxm];int r[maxm];int value[maxm];int len;void dfs(int ind,int indnow){    int v = value[ind];    int vnow = value[indnow];    if(v<vnow)    {        if(l[indnow]==-1)        {            l[indnow] = ind;            return ;        }        else dfs(ind,l[indnow]);    }    else    {        if(r[indnow]==-1)        {            r[indnow] = ind;            return ;        }        else dfs(ind,r[indnow]);    }}void pri(int ind){    if(ind==-1) return ;    if(ind==1) printf("%d",value[ind]);    else printf(" %d",value[ind]);    pri(l[ind]);    pri(r[ind]);}int main() {//    freopen("in.txt","r",stdin);//    srand(3999+time(0));//    printf("%d\n",rand()%5243+1000);    while(scanf("%d",&len)!=EOF)    {        rep1(i,1,len) l[i] = -1,r[i] = -1;        rep1(i,1,len)        {            read1(value[i]);            if(i==1) continue;            dfs(i,1);        }        pri(1);        puts("");    }    return 0;}#define wa  accept#define re  accept#define tle accept#define mle accept#define pe  accept