HDU 5821 ball

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

54 10 0 1 10 1 1 11 44 10 0 1 10 0 2 21 44 21 0 0 00 0 0 11 33 44 21 0 0 00 0 0 13 41 35 21 1 2 2 02 2 1 1 01 32 4

 

Sample Output

NoNoYesNoYes

 

贪心的去想结论，首先两个序列排序完一定是一样的，否则是No

然后考虑b中每个点从哪里来，假如b[i]=k是b中第j个k,那么这个k应该也是a中的第j个k.

于是把a中每个点对应的最终位置标记出来，操作就等于进行区间排序，如果结果不是c[i]=i那就是No了

#include<set>#include<map>#include<ctime>#include<cmath>#include<stack>#include<queue>#include<bitset>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define rep(i,j,k) for (int i = j; i <= k; i++)#define per(i,j,k) for (int i = j; i >= k; i--)#define lson x << 1, l, mid#define rson x << 1 | 1, mid + 1, r#define fi first#define se second#define mp(i,j) make_pair(i,j)#define pii pair<int,int>using namespace std;typedef long long LL;const int low(int x) { return x&-x; }const double eps = 1e-8;const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int N = 1e3 + 10;const int read(){    char ch = getchar();    while (ch<'0' || ch>'9') ch = getchar();    int x = ch - '0';    while ((ch = getchar()) >= '0'&&ch <= '9') x = x * 10 + ch - '0';    return x;}int T, n, m, a[N], b[N], l, r, f[N], c[N], flag;int main(){    T = read();    while (T--)    {        scanf("%d%d", &n, &m);         rep(i, 1, n) a[i] = read();        rep(i, 1, n) b[i] = read();        rep(i, 1, n) f[i] = 0;        flag = 0;        rep(i, 1, n)        {            rep(j, 1, n)            {                if (f[j] || b[i] != a[j]) continue;                flag += f[j] = 1; c[j] = i;  break;            }        }        rep(i, 1, m)        {            scanf("%d%d", &l, &r);            if (flag < n) continue;            sort(c + l, c + r + 1);        }        rep(i, 1, n) if (c[i] != i) flag = 0;        printf("%s\n", flag < n ? "No" : "Yes");    }    return 0;}