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

Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 515    Accepted Submission(s): 309

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

Author

学军中学

Source

2016 Multi-University Training Contest 8

题目大意：给两个长度为n的数组a和b，再给定m次操作，每次给定l和r，每次可以把[l,r]的数进行任意调换位置，问能否在转换后使得a数组变成b数组。

解题思路：用结构体存储a数组，值和下标，下标存的是a数组对应b数组之后的下标。然后在对a数组的下标进行排序，这样就会使a数组的值随着下标的变化而变换，尽可能使a和b相同的值对应好，最后在扫一遍看两个数组相同位置的数字是否相同，相同输出Yes，否则输出No。

详见代码。

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

#define N 1010

struct node
{
    int a,i;//值和下标
} s[N];

int b[N];

bool cmp(node a,node b)
{
    return a.i<b.i;
}

int main()
{
    int t;
    scanf("%d",&t);
    while (t--)
    {
        int n,m,flag=1;
        scanf("%d%d",&n,&m);
        for (int i=0; i<n; i++)
        {
            scanf("%d",&s[i].a);
            s[i].i=-1;
        }
        for (int i=0; i<n; i++)
        {
            scanf("%d",&b[i]);
            for (int j=0; j<n; j++)
            {
                if (s[j].a==b[i]&&s[j].i==-1)
                {
                    s[j].i=i;
                    break;
                }
            }
        }
        int l,r;
        for (int i=0; i<m; i++)
        {
            scanf("%d%d",&l,&r);
            sort(s+l-1,s+r,cmp);//根据s的下标进行排序
        }
        for (int i=0; i<n; i++)
        {
            if (b[i]!=s[i].a)
            {
                flag=0;
                break;
            }
        }
        if (flag==1)
            printf ("Yes\n");
        else
            printf ("No\n");
    }
    return 0;
}