## Friday, August 12, 2016

### HDU 5821 – Ball [思路/贪心] | codekun

http://acm.hdu.edu.cn/showproblem.php?pid=5821
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 1in, 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


HDU 5821 – Ball [思路/贪心] | codekun

### Analysis

typedef pair<int, int> P;

const int maxn = 1024;
P A[maxn];
P seg[maxn];
int N, M, B[maxn];
int cnt[maxn];

bool solve() {
    for(int i = 0; i <= N; ++i)
        if(cnt[i] != 0) return false;
    for(int i = 1; i <= N; ++i) {
        for(int j = 1; j <= N; ++j) {
            if(A[j].first != -1) continue;
            if(A[j].second == B[i]) {
                A[j].first = i;
                break;
            }
        }
    }
    for(int i = 1; i <= M; ++i)
        sort(A + seg[i].first, A + seg[i].second + 1);
    for(int i = 1; i <= N; ++i)
        if(A[i].second != B[i])
            return false;
    return true;
}

int main() {
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &N, &M);
        memset(cnt, 0, sizeof(cnt));
        memset(A, -1, sizeof(A));
        for(int i = 1; i <= N; ++i) {
            scanf("%d", &A[i].second);
            ++cnt[A[i].second];
        }
        for(int i = 1; i <= N; ++i) {
            scanf("%d", B + i);
            --cnt[B[i]];
        }
        for(int i = 1; i <= M; ++i) {
            scanf("%d%d", &seg[i].first, &seg[i].second);
        }
        puts(solve() ? "Yes" : "No");
    }
    return 0;
}