http://acm.hdu.edu.cn/showproblem.php?pid=3410

Passing the Message

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 827    Accepted Submission(s): 546

Problem Description

What
a sunny day! Let’s go picnic and have barbecue! Today, all kids in “Sun
Flower” kindergarten are prepared to have an excursion. Before kicking
off, teacher Liu tells them to stand in a row. Teacher Liu has an
important message to announce, but she doesn’t want to tell them
directly. She just wants the message to spread among the kids by one
telling another. As you know, kids may not retell the message exactly
the same as what they was told, so teacher Liu wants to see how many
versions of message will come out at last. With the result, she can
evaluate the communication skills of those kids.
Because all kids have different height, Teacher Liu set some message passing rules as below:

1.She tells the message to the tallest kid.

2.Every kid who gets the message must retell the message to his “left messenger” and “right messenger”.

3.A kid’s “left messenger” is the kid’s tallest “left follower”.

4.A
kid’s “left follower” is another kid who is on his left, shorter than
him, and can be seen by him. Of course, a kid may have more than one
“left follower”.

5.When a kid looks left, he can only see as far as the nearest kid who is taller than him.

The
definition of “right messenger” is similar to the definition of “left
messenger” except all words “left” should be replaced by words “right”.

For
example, suppose the height of all kids in the row is 4, 1, 6, 3, 5, 2
(in left to right order). In this situation , teacher Liu tells the
message to the 3rd kid, then the 3rd kid passes the message to the 1st
kid who is his “left messenger” and the 5th kid who is his “right
messenger”, and then the 1st kid tells the 2nd kid as well as the 5th
kid tells the 4th kid and the 6th kid.
Your task is just to figure out the message passing route.

Input

The first line contains an integer T indicating the number of test cases, and then T test cases follows.
Each
test case consists of two lines. The first line is an integer N (0< N
<= 50000) which represents the number of kids. The second line lists
the height of all kids, in left to right order. It is guaranteed that
every kid’s height is unique and less than 2^31 – 1 .

Output

For
each test case, print “Case t:” at first ( t is the case No. starting
from 1 ). Then print N lines. The ith line contains two integers which
indicate the position of the ith (i starts form 1 ) kid’s “left
messenger” and “right messenger”. If a kid has no “left messenger” or
“right messenger”, print ‘0’ instead. (The position of the leftmost kid
is 1, and the position of the rightmost kid is N)

Sample Input

2
5
5 2 4 3 1
5
2 1 4 3 5

Sample Output

Case 1:
0 3
0 0
2 4
0 5
0 0
Case 2:
0 2
0 0
1 4
0 0
3 0

Source

2010 National Programming Invitational Contest Host by ZSTU

要找的就是每个元素左右两侧能看见的(如果被更高的挡住就表示看不见)比这个人低的所有人里最高的那个，没有的话输出0；

显然具有单调性，当比栈顶还低/空栈时表示没符合条件的人，输出0并入栈；

当高于栈顶时，由于是一个单调递减栈，一直出栈找到符合条件的最高者记下id即可，记得把这个人最后入栈；

由于i=k时出栈的都是比a[k]矮的人,后面出现a[j]<a[k]的情况显然他看不见pop掉的人，如果a[j]>a[k]则选a[k]显然优于选pop掉的人，所以可以pop掉；

 1 #include <iostream>
 2 #include<algorithm>
 3 #include<stack>
 4 #include<cstdio>
 5 #include<cstring>
 6 using namespace std;
 7 typedef long long LL;
 8 const int MAX = 50005;
 9 int a[MAX], l[MAX], r[MAX];
10 int main()
11 {
12     int N, i, j, k, t;
13     scanf("%d", &t);
14     for (int cas = 1;cas <= t;++cas)
15     {
16         stack<int>S;
17         //memset(l, 0, sizeof(l));
18         //memset(r, 0, sizeof(r));
19         scanf("%d", &N);
20         for (i = 1;i <= N;++i) scanf("%d", &a[i]);
21         for (i = 1;i <= N;++i) {
22             if (S.empty() || a[i]<a[S.top()]) {
23                 l[i] = 0;
24                 S.push(i);
25             }
26             else {
27                 while (!S.empty()&&a[S.top()]<a[i]) {
28                     l[i] = S.top();
29                     S.pop();
30                 }
31                 S.push(i);
32             }
33         }while (!S.empty())S.pop();
34         for (i = N;i >= 1;--i) {
35             if (S.empty() || a[i]<a[S.top()]) {
36                 r[i] = 0;
37                 S.push(i);
38             }
39             else {
40                 while (!S.empty() && a[S.top()]<a[i]) {
41                     r[i] = S.top();
42                     S.pop();
43                 }
44                 S.push(i);
45             }
46         }
47         printf("Case %d:\n", cas);
48         for (i = 1;i <= N;++i) printf("%d %d\n", l[i], r[i]);
49     }
50     return 0;
51 }