（牛客网）字节跳动冬令营网络赛-测试赛 Origami

题目描述

Chiaki has a very big sheet of paper. This sheet has a form of rectangle with dimensions 1 x n and numbers from 1 to n was written on each small 1 x 1 grid. Chiaki would like to fold the paper using the following operations:

• Fold the sheet of paper at position pi to the right. After this query the leftmost part of the paper with dimensions 1 x pi must be above the rightmost part of the paper with dimensions ).
• Fold the sheet of paper at position pi to the left. After this query the rightmost part of the paper with dimensions ) must be above the leftmost part of the paper with dimensions 1 x pi.

After performing the above operations several times, the sheet of paper has dimensions 1 x 1. If we write down the number on each grid from top to bottom, we will have a permutation of n.
Now given a permutation of n, Chiaki would like to know whether it is possible to obtain the permutation using the above operations.

输入描述:

There are multiple test cases. The first line of the input contains an integer T, indicating the number of test cases. For each test case:
The first line contains an integer n (1 ≤ n ≤ 106), indicating the length of the paper.
The second line contains n integers a1, a2, ..., an, which is a permutation of n, indicating the integers marked in the grids of the resulting sheet of paper from top to bottom.
It's guaranteed that the sum of n in all test cases will not exceed 106.

输出描述:

For each test case output one line. If it's possible to obtain the permutation, output ``Yes'' (without the quotes), otherwise output ``No'' (without the quotes).

示例1

输入

3
4
2 1 4 3
7
2 5 4 3 6 1 7
4
1 3 2 4

输出

Yes
Yes
No

思路：分为左右两边连线，连线不相交就成立。

#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<stack>
#include<vector>
#define M 1000010
using namespace std;
struct node
{
    int x,y;
    node() {};
    node(int a,int b)
    {
        x=min(a,b);
        y=max(a,b);
    }
};
vector<node>L,R;
int t,n,x;
int A[M],B[M];
int pan(vector<node>&v)
{
    stack<int>q;
    memset(B,0,4*(2+n));//改成menset(B,0,sizeof(B))会超时
    int len=v.size();
    for(int i=0; i<len; i++)
    {
        B[v[i].x]=i+1;
        B[v[i].y]=-i-1;
    }
    for(int i=1; i<=n; i++)
    {
        if(B[i]>0)
            q.push(B[i]);
        else if(B[i]<0)
        {
            if(q.top()+B[i]==0)
                q.pop();
            else
                return false;
        }
    }
    return true;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        L.clear();
        R.clear();
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&x);
            A[x]=i;
        }
        for(int i=1; i<n; i++)
        {
            node w(A[i],A[i+1]);
            if(i&1) L.push_back(w);
            else R.push_back(w);
        }
        if(pan(L)&&pan(R))
            printf("Yes\n");
        else
            printf("No\n");
    }
}