hdu3410-Passing the Message(RMQ,感觉我写的有点多此一举。。。其实可以用单调栈)

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

 

题意：简单点说就是找到每个数他左边比它小的最大的数以及右边比它小的最大的数，但是有限制条件，以左边为例，只能在左边第一个比它大的数（如果有的话）和它之间

的范围里找，右边同理。

 

解析：有单调栈的解法，我这里说一下RMQ如何做，我在数组最左边和最右边添加一个很大的数(确保任何一个数都可以确定范围)，然后以某个位置i为例，我用RMQ查找0到

i-1之间第一个比它大的数（二分+RMQ），找到位置j,再用RMQ直接找j+1到i-1之间最大值（如果j+1==i,则没有）。

 

代码

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<utility>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<iterator>
#include<stack>
using namespace std;
const int INF=1e9+;
const double eps=1e-;
const int maxn=;
int N,A[maxn],B[maxn];
int Lg[maxn],ansl[maxn],ansr[maxn];
void GetLg()
{
    Lg[]=-;
    for(int i=;i<maxn;i++) Lg[i]=Lg[i-]+((i&(i-))?:);
}
typedef pair<int,int> par;
par rmq[][maxn][];
int f(int x){ return <<x; }
void ST()   //保存最大值，同时把值和下标保存下来
{
    for(int i=;i<=N+;i++) rmq[][i][]=make_pair(A[i],i); //下标为正是为了方便找左边第一个比某个数大的位置
    for(int i=;i<=N+;i++) rmq[][i][]=make_pair(A[i],-i); //下标为负是为了方便找右边第一个比某个数大的位置
    for(int k=;k<;k++)
    for(int j=;f(j)<=N+;j++)
        for(int i=;i+f(j)-<=N+;i++)
           rmq[k][i][j]=max(rmq[k][i][j-],rmq[k][i+f(j-)][j-]);
}
par RMQ(int x,int y,int type)
{
    if(x>y) swap(x,y);
    int k=Lg[y-x+];
    return max(rmq[type][x][k],rmq[type][y-f(k)+][k]);
}
int GetPos(int x,int y,int type,int v)
{
    while(true)
    {
        if(x==y) return x;
        int mid=(x+y)/;
        par a=RMQ(x,mid,type);
        par b=RMQ(mid+,y,type);
        if(type==)
        {
            if(b.first>v) x=mid+;
            else y=mid;
        }
        if(type==)
        {
            if(a.first>v) y=mid;
            else x=mid+;
        }
    }
}
int main()
{
    GetLg();
    int T,Case=;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        A[]=A[N+]=INF;
        for(int i=;i<=N;i++)
        {
            scanf("%d",&A[i]);
            B[i]=A[i];
        }
        sort(B+,B+N+);
        for(int i=;i<=N;i++) A[i]=lower_bound(B+,B+N+,A[i])-B;//离散化
        ST();
        for(int i=;i<=N;i++)
        {
            int pos=GetPos(,i-,,A[i]);  //找到左边第一个比他大的数的位置
            if(pos+==i) ansl[i]=;      //中间没有比他小的数
            else  ansl[i]=RMQ(pos+,i-,).second;
        }
        for(int i=N;i>=;i--)
        {
            int pos=GetPos(i+,N+,,A[i]);
            if(pos-==i) ansr[i]=;
            else  ansr[i]=-RMQ(i+,pos-,).second;
        }
        printf("Case %d:\n",++Case);
        for(int i=;i<=N;i++) printf("%d %d\n",ansl[i],ansr[i]);
    }
    return ;
}