先上题目：
Binary Search Heap Construction
Time Limit: 5 Seconds Memory Limit: 32768 KB
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/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：
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 }
/*单调栈*/
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 }
/*RMQ*/
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