Look Up

Problem 1845 Look Up

Accept: 173    Submit: 543
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

N (1 <= N <= 100,000) monkeys in the mountains, conveniently numbered 1..N, are once again standing in a row. Monkey i has height H_i (1 <= H_i <= 1,000,000).

Each monkey is looking to his left toward those with higher index numbers. We say that monkey i "looks up" to monkey j if i < j and H_i < H_j. For each monkey i, we would like to know the index of the first monkey in line looked up to by monkey i.

 Input

Input consists of several testcases. The format of each case as follow:

• Line 1: A single integer: N
• Lines 2..N+1: Line i+1 contains the single integer: H_i

 

 Output

For each testcase, output N lines. Line i contains a single integer representing the smallest index of a monkey up to which monkey i looks. If no such monkey exists, print 0.

 Sample Input

6 3 2 6 1 1 2

 Sample Output

3 3 0 6 6 0

 Hint

Monkey 1 and 2 both look up to monkey 3; monkey 4 and 5 both look up to monkey 6; and monkey 3 and 6 do not look up to any monkey.

 Source

Funny Programming Contest -- OSUM

 

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用优先队列保存输入信息。

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <stack>
 4 #include <queue>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <cstring>
 8 using namespace std;
 9 
10 struct Height
11 {
12     int h, id;
13 }hei[100005];
14 int ans[100005], n;
15 priority_queue<Height> Q;
16 
17 bool operator < (const Height& x, const Height& y)
18 {
19     return x.h > y.h;
20 }
21 
22 int main()
23 {
24     while(scanf("%d", &n) != EOF)
25     {
26         for(int i = 0; i < n; i++)
27         {
28             scanf("%d", &hei[i].h);
29             hei[i].id = i;
30             while(!Q.empty() && hei[i].h > (Q.top()).h)
31             {
32                 ans[(Q.top()).id] = i + 1;
33                 Q.pop();
34             }
35             Q.push(hei[i]);
36         }
37         while(!Q.empty())
38         {
39             ans[(Q.top()).id] = 0;
40             Q.pop();
41         }
42         for(int i = 0; i < n; i++) printf("%d\n", ans[i]);
43     }
44     return 0;
45 }