# Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 854    Accepted Submission(s): 502

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

Solution

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ms(a) memset(a,0,sizeof(a))
typedef long long ll;
const int INF = INT_MAX;
const int MAXN = 100000 + 10;
using namespace std;
struct node
{
int num, dir;
bool operator < (const node b)const
{
return dir < b.dir;
}
}t;
int main()
{
int T;
cin >> T;
while (T--)
{
vector<node> v;
vector<int> b;
int n, m;
cin >> n >> m;
for (int i = 0; i < n; i++)
{
scanf("%d", &t.num); t.dir = -1;
v.push_back(t);
}
for (int i = 0; i < n; i++)
{
int t; scanf("%d", &t);
b.push_back(t);
for (int j = 0; j < n; j++)
{
if (b[i] == v[j].num&&v[j].dir == -1)
{
v[j].dir = i; break;
}
}
}
for (int i = 0; i < m; i++)
{
int l, r; scanf("%d%d", &l, &r);
l--;//sort的参数2是要加1的，所以r就不减了。
sort(v.begin() + l, v.begin() + r);
}
bool f = 1;
for (int i = 0; i < n; i++)
if (v[i].num != b[i])
{
f = 0; break;
}
if (f)printf("Yes\n");
else printf("No\n");
}
return 0;
}