HDU-5821 Ball（排序）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5821点击打开链接

Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1378    Accepted Submission(s): 781

Problem Description

ZZX has a sequence of boxes numbered  1,2,...,n. Each box can contain at most one ball.

You are given the initial configuration of the balls. For  1≤i≤n, if the  i-th box is empty then  a[i]=0, otherwise the i-th box contains exactly one ball, the color of which is a[i], a positive integer. Balls with the same color cannot be distinguished.

He will perform m operations in order. At the i-th operation, he collects all the balls from boxes l[i],l[i]+1,...,r[i]-1,r[i], and then arbitrarily put them back to these boxes. (Note that each box should always contain at most one ball)

He wants to change the configuration of the balls from a[1..n] to b[1..n] (given in the same format as a[1..n]), using these operations. Please tell him whether it is possible to achieve his goal.

 

Input

First line contains an integer t. Then t testcases follow. 
In each testcase: First line contains two integers n and m. Second line contains a[1],a[2],...,a[n]. Third line contains b[1],b[2],...,b[n]. Each of the next m lines contains two integers l[i],r[i].

1<=n<=1000,0<=m<=1000, sum of n over all testcases <=2000, sum of m over all testcases <=2000.

0<=a[i],b[i]<=n.

1<=l[i]<=r[i]<=n.

 

Output

For each testcase, print "Yes" or "No" in a line.

 

Sample Input

5 4 1 0 0 1 1 0 1 1 1 1 4 4 1 0 0 1 1 0 0 2 2 1 4 4 2 1 0 0 0 0 0 0 1 1 3 3 4 4 2 1 0 0 0 0 0 0 1 3 4 1 3 5 2 1 1 2 2 0 2 2 1 1 0 1 3 2 4

 

Sample Output

No No Yes No Yes

 

将b数组看做排序后的结果 然后每次操作就是对区间排序

如此一来便可以将a数组通过下标转化成乱序的数组 对每次操作进行桶排序 最后检验是否一致即可

注意构造下标时因为不会出现交叉情况 因此重复数字对其没有影响

这里因为数的范围时0~n 会少了离散操作

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <set>
using namespace std;
int a[2222];
int b[2222];
int  main()
{
        int t;
        scanf("%d",&t);
        while(t--)
        {
                int n,m;
                scanf("%d%d",&n,&m);
                for(int i=0;i<n;i++)
                        scanf("%d",&a[i]);
                for(int i=0;i<n;i++)
                        scanf("%d",&b[i]);
                vector<int > pos[n+1];
                int point[n+1];
                int num[n+1];
                int nmid=n+1;
                for(int i=0;i<=n;i++)
                        point[i]=0;
                for(int i=0;i<n;i++)
                        {
                            pos[b[i]].push_back(i);
                        }
                for(int i=0;i<=n;i++)
                        num[i]=pos[i].size();
                for(int i=0;i<n;i++)
                {
                        if(point[a[i]]<num[a[i]])
                        {
                                int mid=a[i];
                                a[i]=pos[a[i]][point[a[i]]];
                                point[mid]++;
                        }
                        else
                                a[i]=nmid++;
                }
                for(int i=0;i<m;i++)
                {
                        int l,r;
                        scanf("%d%d",&l,&r);
                        int tong[2222];
                        for(int i=0;i<2222;i++)
                                tong[i]=0;
                        for(int i=l-1;i<=r-1;i++)
                        {
                                tong[a[i]]++;
                        }
                        int cnt=0;
                        while(!tong[cnt])
                               cnt++;
                        for(int i=l-1;i<=r-1;i++)
                        {
                                a[i]=cnt;
                                tong[cnt]--;
                                while(!tong[cnt])
                                        cnt++;
                        }

                }
                int flag=0;
                for(int i=0;i<n;i++)
                        if(a[i]!=i)
                {
                        flag=1;
                }
                if(flag)
                        cout << "No" << endl;
                else
                        cout << "Yes" << endl;
        }
}