hdu 5821 Ball (贪心)

Ball

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 \leq i \leq 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

贪心的策略是让A数组中的数更接近正确位置，正确位置是A数组中的数在B数组中出现的第一个未被标记的位置，让数接近正确位置的贪心就是对于每个区间sort。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1010;
int a[maxn], b[maxn];
int id[maxn];
int main() {
    int t;
    scanf("%d", &t);
    while(t--) {
        int n, m;
        scanf("%d %d", &n, &m);
        memset(id, 0, sizeof(id));
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        for(int i = 1; i <= n; i++) {
            scanf("%d", &b[i]);
            for(int j = 1; j <= n; j++) {
                if(!id[j] && a[j] == b[i]) {
                    id[j] = i;
                    break;
                }
            }
        }
        for(int i = 1; i <= m; i++) {
            int l, r;
            scanf("%d %d", &l, &r);
            sort(id + l, id + r + 1);
        }
        int flag = 1;
        for(int i = 1; i <= n; i++) {
            if(id[i] != i) {
                flag = 0;
                break;
            }
        }
        if(flag) puts("Yes");
        else puts("No");
    }
}