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[200];
    int p;

    bool operator < (const node& o)const{
        return strcmp(l,o.l)<0;
    }
}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=0;i<ti;i++) putchar(c);
}

int main()
{
    char* sp;
    //freopen("data.txt","r",stdin);
    while(scanf("%d",&n),n){
        for(int i=1;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]=0;
        }
        sort(e+1,e+n+1);
        e[0].p=e[n+1].p=INF;
        for(int i=1;i<=n;i++){
            while(e[i].p>=e[l[i]-1].p) l[i]=l[l[i]-1];
            al[l[i]]++;
        }
        for(int i=n;i>0;i--){
            while(e[i].p>=e[r[i]+1].p) r[i]=r[r[i]+1];
            ar[r[i]]++;
        }
        for(int i=1;i<=n;i++){
            put('(',al[i]);
            printf("%s/%d",e[i].l,e[i].p);
            put(')',ar[i]);
        }
        printf("\n");
    }
    return 0;
}

#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 1;
        return 0;
    }
}node;

int n;
node e[MAX];
char ss[MAX];
int dp[MAX][20];

void solve(){
    int i,j,l,r;
    for(i=0;i<n;i++) dp[i][0]=i;
    for(j=1;(1<<j)<=n;j++){
        for(i=0;i+(1<<j)-1<n;i++){
            l=dp[i][j-1];   r=dp[i+(1<<(j-1))][j-1];
            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-(1<<k)+1][k]].p ? dp[a][k] : dp[b-(1<<k)+1][k]);
}

void print(int r,int L,int R){
    int ne;
    putchar('(');
    if(r-L>0){
        ne=rmq(L,r-1);
        print(ne,L,r-1);
    }
    for(unsigned int i=0;i<e[r].l.size();i++) putchar(e[r].l[i]);
    printf("/%lld",e[r].p);
    if(R-r>0){
        ne=rmq(r+1,R);
        print(ne,r+1,R);
    }
    putchar(')');
}

int main()
{
    int li;
    char* st;
    //freopen("data.txt","r",stdin);
    while(scanf("%d",&n),n!=0){
        for(int i=0;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=0;
        for(int i=1;i<n;i++){
            if(e[li].p<e[i].p) li=i;
        }
        print(li,0,n-1);
        printf("\n");
    }
    return 0;
}