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ZOJ - 2243 - Binary Search Heap Construction
先上题目：
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,...,ln/pndenoting 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/17 a/1 b/2 c/3 d/4 e/5 f/6 g/77 a/3 b/6 c/4 d/7 e/2 f/5 g/10
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：
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define MAX 50002 5 #define INF 0x3f3f3f3f 6 using namespace std; 7 8 typedef struct node{ 9 char l[200];10 int p;11 12 bool operator < (const node& o)const{13 return strcmp(l,o.l)<0;14 }15 }node;16 17 int n;18 node e[MAX];19 char ch[MAX];20 int l[MAX],r[MAX];21 int al[MAX],ar[MAX];22 23 inline void put(char c,int ti){24 for(int i=0;i<ti;i++) putchar(c);25 }26 27 int main()28 {29 char* sp;30 //freopen("data.txt","r",stdin);31 while(scanf("%d",&n),n){32 for(int i=1;i<=n;i++){33 scanf("%s",ch);34 sp=strchr(ch,‘/‘);35 *sp=‘\0‘;36 sp++;37 strcpy(e[i].l,ch);38 sscanf(sp,"%d",&e[i].p);39 l[i]=r[i]=i;40 al[i]=ar[i]=0;41 }42 sort(e+1,e+n+1);43 e[0].p=e[n+1].p=INF;44 for(int i=1;i<=n;i++){45 while(e[i].p>=e[l[i]-1].p) l[i]=l[l[i]-1];46 al[l[i]]++;47 }48 for(int i=n;i>0;i--){49 while(e[i].p>=e[r[i]+1].p) r[i]=r[r[i]+1];50 ar[r[i]]++;51 }52 for(int i=1;i<=n;i++){53 put(‘(‘,al[i]);54 printf("%s/%d",e[i].l,e[i].p);55 put(‘)‘,ar[i]);56 }57 printf("\n");58 }59 return 0;60 }
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <string> 5 #include <cmath> 6 #define MAX 60002 7 #define ll long long 8 using namespace std; 9 10 typedef struct node{11 string l;12 ll p;13 14 bool operator <(const node& o)const{15 if(l<o.l) return 1;16 return 0;17 }18 }node;19 20 int n;21 node e[MAX];22 char ss[MAX];23 int dp[MAX][20];24 25 void solve(){26 int i,j,l,r;27 for(i=0;i<n;i++) dp[i][0]=i;28 for(j=1;(1<<j)<=n;j++){29 for(i=0;i+(1<<j)-1<n;i++){30 l=dp[i][j-1]; r=dp[i+(1<<(j-1))][j-1];31 if(e[l].p>e[r].p) dp[i][j]=l;32 else dp[i][j]=r;33 }34 }35 }36 37 int rmq(int a,int b){38 int k;39 k=log(b-a+1.0)/log(2.0);40 return (e[dp[a][k]].p>e[dp[b-(1<<k)+1][k]].p ? dp[a][k] : dp[b-(1<<k)+1][k]);41 }42 43 void print(int r,int L,int R){44 int ne;45 putchar(‘(‘);46 if(r-L>0){47 ne=rmq(L,r-1);48 print(ne,L,r-1);49 }50 for(unsigned int i=0;i<e[r].l.size();i++) putchar(e[r].l[i]);51 printf("/%lld",e[r].p);52 if(R-r>0){53 ne=rmq(r+1,R);54 print(ne,r+1,R);55 }56 putchar(‘)‘);57 }58 59 int main()60 {61 int li;62 char* st;63 //freopen("data.txt","r",stdin);64 while(scanf("%d",&n),n!=0){65 for(int i=0;i<n;i++){66 getchar();67 scanf("%s",ss);68 st=strchr(ss,‘/‘);69 *st=‘\0‘;70 st++;71 e[i].l=string(ss);72 sscanf(st,"%lld",&e[i].p);73 }74 sort(e,e+n);75 solve();76 li=0;77 for(int i=1;i<n;i++){78 if(e[li].p<e[i].p) li=i;79 }80 print(li,0,n-1);81 printf("\n");82 }83 return 0;84 }