ZOJ - 2243 - Binary Search Heap Construction
2014-08-11 21:03
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先上题目:
Binary Search Heap ConstructionTime 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
Sample Output
/*RMQ*/
Binary Search Heap ConstructionTime 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:
#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; }
/*RMQ*/
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