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 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

HDU 5821 – Ball [思路/贪心] | codekun
给定两个长度为N的数列A和B，有M次操作，每次可以把Li到Ri段的数字打乱重排，问你能否通过M次操作把A变成B

Analysis

首先能够想到的是如果数列B中有3个2，如果能把A变成B，那么可以定为3个2之间是不交叉交换的。那么这样根据最终的序列就有一个优先级，最后越靠左的数在重排时越靠左，这样我们根据优先级对A序列编号后排序M次，看最终能否变成B即可。

正解应该是贪心，对于最左边的值，那么他一定对应着在B[i]数组中与他相同的，且最左边的数。

根据这个，贪心的去扫一遍就好了，其实就是不断排序，为什么？

因为排序之后，一定会使得从目标到结束状态花费变得最小。

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;

}

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