Passing the Message 单调队列

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

#include <iostream>
#include <cstdio>
#include <queue>
#include <string>
#include <cstring>
using namespace std;

#define rep(x,y) for(int i=x;i<=y;i++)
#define repp(x,y,m,n) rep(x,m)rep(y,n)

typedef long long ll;


const int M=50000+5;
int a[M];
int deq[M];
int r[M];
int l[M];

void solve(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        printf("Case %d:\n",i);
        int m;
        scanf("%d",&m);
        memset(r,0,sizeof(r));
        memset(l,0,sizeof(l));
        for(int j=1;j<=m;j++){
            scanf("%d",&a[j]);
        }

        int t=0;
        for(int j=1;j<=m;j++){
            while(t!=0&&a[deq[t]]<a[j]){
                --t;
            }
            deq[++t]=j;

            r[deq[t-1]]=deq[t];
        }
        t=0;
        for(int j=m;j>=1;j--){
            while(t!=0&&a[deq[t]]<a[j]){
                --t;
            }
            deq[++t]=j;

            l[deq[t-1]]=deq[t];
        }

        for(int i=1;i<=m;i++){
            printf("%d %d\n",l[i],r[i]);
        }
    }

    return;
}


int main(){
    solve();
    return 0;
}





### OTFS 中的消息传递机制 在正交时间频率空间 (OTFS) 调制系统中,消息传递算法被广泛应用于信道估计和稀疏信号恢复。这类算法通过迭代的方式更新节点之间的信息交换来优化解码过程。 #### 消息传递的工作原理 消息传递基于因子图模型来进行计算。具体来说,在OTFS通信环境中,接收端会构建一个表示信道特性的因子图。该图由变量节点和因子节点组成: - **变量节点**代表待估参数(如信道系数) - **因子节点**则对应于观测数据与这些未知数之间关系的约束条件 每一轮迭代过程中,各节点沿边向相邻节点发送消息,这些消息包含了当前对于关联参数的最佳猜测以及不确定性度量。随着迭代次数增加,整个网络逐渐收敛到最优解附近[^2]。 #### 实现细节 为了更好地理解这一过程,下面给出一段简化版Python伪代码展示如何在一个理想化的场景下执行基本的消息传递操作: ```python import numpy as np def message_passing(observations, factor_graph): num_iterations = 10 # 初始化所有变量节点的状态 var_states = initialize_variable_nodes(factor_graph) for _ in range(num_iterations): new_var_states = {} # 更新每个变量节点状态 for node_id in factor_graph.variables: incoming_messages = collect_incoming_messages(node_id, factor_graph, var_states) updated_state = update_node_state(incoming_messages) new_var_states[node_id] = updated_state var_states = new_var_states return extract_estimates(var_states) def initialize_variable_nodes(graph): initial_guesses = {} for v in graph.variables: # 基于先验知识设定初始值 guess = generate_initial_guess(v.prior_info) initial_guesses[v.id] = guess return initial_guesses def collect_incoming_messages(target_node_id, graph, current_states): messages = [] neighbors = get_neighbor_factors(target_node_id, graph) for f in neighbors: msg_from_f_to_v = compute_message(f, target_node_id, current_states) messages.append(msg_from_f_to_v) return messages def update_node_state(messages): combined_msg = combine_messages(messages) posterior_distribution = apply_bayesian_update(combined_msg) sample = draw_sample(posterior_distribution) uncertainty = estimate_uncertainty(posterior_distribution) return {'value': sample, 'uncertainty': uncertainty} ``` 此段代码展示了消息传递的核心流程:初始化、收集来自邻居的信息并据此调整局部信念直到达到稳定点或最大迭代次数为止。
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