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Description of the topic

Well-known foodie Xiao A was invited to the ATM Hotel to taste the dishes. ATM hotel is small $A$A get ready $N$N The dishes are ordered in the order of 1 to N according to the quality of the dishes. The highest quality dishes are estimated to be $1$1。

Due to the taste of the dishes, some dishes must be made before other dishes, specifically $M$M Strip as " $i$i No. 'required' prior to $j$j The limitation of the production of dishes, we will abbreviate this limitation as $&lt;i,j&gt;$<i,j>。

Now, the hotel hopes to find the order in which an optimal dish is made, making it small $A$ATry to eat high quality dishes first:

That is,

(1) Subject to all restrictions, $1$1 No. dishes "as far as possible" priority production;

(2) in meeting all restrictions, $1$1No. dishes "as far as possible" under the premise of making $2$2No. dishes "as far as possible" priority production;

(3) in meeting all restrictions, $1$1Number and $2$2No. dishes "as far as possible", $3$3No. dishes "as far as possible" priority production;

(4) in meeting all restrictions, $1$1 Number and $2$2 Number and $3$3 No. dishes "as far as possible", $4$4 No. dishes "as far as possible" priority production;

(5) and so on.

Example 1: Total $4$4 Road dishes, two restrictions $&lt;3,1&gt;、&lt;4,1&gt;$<3,1>、<4,1>, then the order of production is $3,4,1,2$3,4,1,2。

Example 2: Total $5$5 Road dishes, two restrictions $&lt;5,2&gt;、 &lt;4,3&gt;$<5,2>、<4,3>, then the order of production is $1,5,2,4,3$1,5,2,4,3。

In Example 1, first consider $1$1Because there are restrictions $&lt;3,1&gt;$<3,1>with $&lt;4,1&gt;$<4,1>So only finished $3$3 with $4$4 After making $1$1And according to $(3)$(3)， $3$3 The number should be as "as far as possible" $4$4 Priority is given, so it is currently determined that the order of the first three dishes is $3,4,1$3,4,1Next consideration $2$2To determine the final production order is $3,4,1,2$3,4,1,2。

In Example 2, first make $1$1 Is not against the restrictions; next consider $2$2 When $&lt;5,2&gt;$<5,2>Limit, so make it first $5$5 Reproduction $2$2Next consideration $3$3 When $&lt;4,3&gt;$<4,3>Limit, so make it first $4$4 Reproduction $3$3, so the final order is $1,5,2,4,3$1,5,2,4,3. Now you need to find the best order for cooking this dish. No solution output "Impossible!" (without quotes, initial capitalization, lowercase letters)

Input and output format

Input format:

The first line is a positive integer $D$D, indicating the number of data sets. Next is $D$DGroup data. For each set of data: the first line of two positive integers separated by spaces $N$Nwith $M$M, the number of entries indicating the number of dishes and the order of production. Next $M$MRow, two positive integers per line $x,y$x,y, indicating" $x$xNo. dishes must be made before the y dishes. (Note: There may be identical restrictions in the M restrictions)

Output format:

Output file only contains $D$D Line, each line $N$N An integer that indicates the optimal order in which the dishes are made, or "Impossible!" means no solution (without quotes).

Input and output sample

Enter sample #1:

3
5 4
5 4
5 3
4 2
3 2
3 3
1 2
2 3
3 1
5 2
5 2
4 3

Output sample #1:

1 5 3 4 2 
Impossible! 
1 5 2 4 3

Description

[example explanation]

The second set of data also requires dishes $1$1Prior to the dishes $2$2Production, dishes $2$2Prior to the dishes $3$3Production, dishes $3$3Prior to the dishes $1$1Production, and this is not possible in any case, resulting in no solution.

100% of the data is met $N,M\le 100000,D\le 3$N,M≤100000,D≤3。

Problem analysis

The problem is trying to make us believe that this is a sort of topological and lexicographical order, but this is not the third example...

Because we are greedy to make the lexicographical order small in front, but what we want is to try to put the smallest in front as far as possible, so obviously $WA$WAof.

Then we consider it in turn, try to put the smallest in front as far as possible, just try to put the maximum in the back, just do it again to make the lexicographically largest solution, and then reverse the output.

code show as below:

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#define R register
#define IN inline
#define W while
#define gc getchar()
#define MX 100500
template <class T>
IN void in(T &x)
{
    x = 0; R char c = gc;
    for (; !isdigit(c); c = gc);
    for (;  isdigit(c); c = gc)
    x = (x << 1) + (x << 3) + c - 48;
}
int n, m;
std::vector <int> nex[MX];
std::priority_queue <int> pq;
int deg[MX], res[MX];
IN bool cmp(R int x, R int y) {return x > y;}
int main(void)
{
    int T, a, b, tot, now;
    in(T);
    W (T--)
    {
        in(n), in(m), tot = 0;
        std::memset(deg, 0, sizeof(deg));
        for (R int i = 1; i <= n; ++i) nex[i].clear();
        for (R int i = 1; i <= m; ++i)
        {
            in(a), in(b); deg[a]++;
            nex[b].push_back(a);
        }
        for (R int i = n; i; --i)
        {
            if (!deg[i]) pq.push(i);
            std::sort(nex[i].begin(), nex[i].end(), cmp);
        }
        W (!pq.empty())
        {
            now = pq.top(), pq.pop();
            res[++tot] = now;
            for (auto i : nex[now])
            {
                --deg[i];
                if (!deg[i]) pq.push(i);
            }
        }
        if (tot != n) puts("Impossible!");
        else
        {
            for (R int i = n; i; --i) printf("%d ", res[i]);
            puts("");
        }
    }
}