ZOJ - 2243 - Binary Search Heap Construction

先上题目：

Binary Search Heap Construction

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Time Limit: 5 Seconds      Memory Limit: 32768 KB

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Read the statement of problem G for the definitions concerning trees. In the following we define the basic terminology of heaps. A heap is a tree whose internal nodes have each assigned a priority (a number) such that the priority of each internal node is less than the priority of its parent. As a consequence, the root has the greatest priority in the tree, which is one of the reasons why heaps can be used for the implementation of priority queues and for sorting.

A binary tree in which each internal node has both a label and a priority, and which is both a binary search tree with respect to the labels and a heap with respect to the priorities, is called a treap. Your task is, given a set of label-priority-pairs, with unique labels and unique priorities, to construct a treap containing this data.

Input Specification

The input contains several test cases. Every test case starts with an integer n. You may assume that 1<=n<=50000. Then follow n pairs of strings and numbers l₁/p₁,...,l_n/p_ndenoting the label and priority of each node. The strings are non-empty and composed of lower-case letters, and the numbers are non-negative integers. The last test case is followed by a zero.

Output Specification

For each test case output on a single line a treap that contains the specified nodes. A treap is printed as (<left sub-treap><label>/<priority><right sub-treap>). The sub-treaps are printed recursively, and omitted if leafs.

Sample Input

7 a/7 b/6 c/5 d/4 e/3 f/2 g/1
7 a/1 b/2 c/3 d/4 e/5 f/6 g/7
7 a/3 b/6 c/4 d/7 e/2 f/5 g/1
0

Sample Output

(a/7(b/6(c/5(d/4(e/3(f/2(g/1)))))))
(((((((a/1)b/2)c/3)d/4)e/5)f/6)g/7)
(((a/3)b/6(c/4))d/7((e/2)f/5(g/1)))

　　题意：好像就是叫你求Treap树。给出字符串和优先值，要求建一棵二叉树，根据字符串排序，然后父亲的优先值要比儿子大。然后先序遍历输出这个Treap树。
　　最水的方法就是直接先按优先值排序，然后逐个逐个元素添加。但是这样做绝对超时。
　　可以通过的第一种方法：首先先按字符串大小排个序，然后从小到大扫描一次，求出每个元素左边优先值比它大的最近的元素的位置在哪。同理从大到小扫描，求出每个元素右边优先值比它大的最近的元素的位置在哪。(没错，就是单调栈),然后在每个扫描到的最近位置加一个对应的括号(左括号或者右括号)就是答案了。总的时间复杂度O(nlogn)。
　　第二种方法是用RMQ求出区间优先值的最大值的下标，然后每次找出区间最大值作为根构造两边的子树就可以了。总的时间复杂度也是O(nlogn)。

　　比赛的时候用的方法是第二种，但是当时求对数的时候底数不是2，所以提交一直都是段错误。

上代码：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #define MAX 50002
 #define INF 0x3f3f3f3f
 using namespace std;

 typedef struct node{
     char l[];
     int p;

     bool operator < (const node& o)const{
         return strcmp(l,o.l)<;
     }
 }node;

 int n;
 node e[MAX];
 char ch[MAX];
 int l[MAX],r[MAX];
 int al[MAX],ar[MAX];

 inline void put(char c,int ti){
     for(int i=;i<ti;i++) putchar(c);
 }

 int main()
 {
     char* sp;
     //freopen("data.txt","r",stdin);
     while(scanf("%d",&n),n){
         for(int i=;i<=n;i++){
             scanf("%s",ch);
             sp=strchr(ch,'/');
             *sp='\0';
             sp++;
             strcpy(e[i].l,ch);
             sscanf(sp,"%d",&e[i].p);
             l[i]=r[i]=i;
             al[i]=ar[i]=;
         }
         sort(e+,e+n+);
         e[].p=e[n+].p=INF;
         for(int i=;i<=n;i++){
             while(e[i].p>=e[l[i]-].p) l[i]=l[l[i]-];
             al[l[i]]++;
         }
         for(int i=n;i>;i--){
             while(e[i].p>=e[r[i]+].p) r[i]=r[r[i]+];
             ar[r[i]]++;
         }
         for(int i=;i<=n;i++){
             put('(',al[i]);
             printf("%s/%d",e[i].l,e[i].p);
             put(')',ar[i]);
         }
         printf("\n");
     }
     return ;
 }

/*单调栈*/

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <string>
 #include <cmath>
 #define MAX 60002
 #define ll long long
 using namespace std;

 typedef struct node{
     string l;
     ll p;

     bool operator <(const node& o)const{
         if(l<o.l) return ;
         return ;
     }
 }node;

 int n;
 node e[MAX];
 char ss[MAX];
 int dp[MAX][];

 void solve(){
     int i,j,l,r;
     for(i=;i<n;i++) dp[i][]=i;
     for(j=;(<<j)<=n;j++){
         for(i=;i+(<<j)-<n;i++){
             l=dp[i][j-];   r=dp[i+(<<(j-))][j-];
             if(e[l].p>e[r].p) dp[i][j]=l;
             else dp[i][j]=r;
         }
     }
 }

 int rmq(int a,int b){
     int k;
     k=log(b-a+1.0)/log(2.0);
     return (e[dp[a][k]].p>e[dp[b-(<<k)+][k]].p ? dp[a][k] : dp[b-(<<k)+][k]);
 }

 void print(int r,int L,int R){
     int ne;
     putchar('(');
     if(r-L>){
         ne=rmq(L,r-);
         print(ne,L,r-);
     }
     for(unsigned int i=;i<e[r].l.size();i++) putchar(e[r].l[i]);
     printf("/%lld",e[r].p);
     if(R-r>){
         ne=rmq(r+,R);
         print(ne,r+,R);
     }
     putchar(')');
 }

 int main()
 {
     int li;
     char* st;
     //freopen("data.txt","r",stdin);
     while(scanf("%d",&n),n!=){
         for(int i=;i<n;i++){
             getchar();
             scanf("%s",ss);
             st=strchr(ss,'/');
             *st='\0';
             st++;
             e[i].l=string(ss);
             sscanf(st,"%lld",&e[i].p);
         }
         sort(e,e+n);
         solve();
         li=;
         for(int i=;i<n;i++){
             if(e[li].p<e[i].p) li=i;
         }
         print(li,,n-);
         printf("\n");
     }
     return ;
 }

/*RMQ*/