18年春季第三题 PAT甲级 1146 Topological Order (25分)

谁践踏了优雅 2023-07-14 10:38 4阅读 0赞

题目来源：[https://pintia.cn/problem-sets/994805342720868352/problems/994805343043829760][https_pintia.cn_problem-sets_994805342720868352_problems_994805343043829760]

备考汇总贴：[2020年6月PAT甲级满分必备刷题技巧][2020_6_PAT]

This is a problem given in the Graduate Entrance Exam in 2018: Which of the following is NOT a topological order obtained from the given directed graph? Now you are supposed to write a program to test each of the options.

![gre.jpg][]

### Input Specification: ###

Each input file contains one test case. For each case, the first line gives two positive integers N (≤ 1,000), the number of vertices in the graph, and M (≤ 10,000), the number of directed edges. Then M lines follow, each gives the start and the end vertices of an edge. The vertices are numbered from 1 to N. After the graph, there is another positive integer K (≤ 100). Then K lines of query follow, each gives a permutation of all the vertices. All the numbers in a line are separated by a space.

### Output Specification: ###

Print in a line all the indices of queries which correspond to "NOT a topological order". The indices start from zero. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line. It is graranteed that there is at least one answer.

### Sample Input: ###

6 8
    1 2
    1 3
    5 2
    5 4
    2 3
    2 6
    3 4
    6 4
    5
    1 5 2 3 6 4
    5 1 2 6 3 4
    5 1 2 3 6 4
    5 2 1 6 3 4
    1 2 3 4 5 6

### Sample Output: ###

3 4

# 题意（源自[liuchuo][]，经过了我的编辑补充） #

给一个有向图，判断给定序列是否是拓扑序列。

# 分析 #

1.拓扑序列在近三年考题中只出现了一次，有可能再考。由于关于拓扑的练习题在甲级题库里只有一道，补充PTA的另一道题：[任务调度的合理性 (25分)][_25]

2.用邻接表v存储这个有向图，并将每个节点的入度保存在in数组中。对每一个要判断是否是拓扑序列的结点遍历，如果当前结点的入度不为0则表示不是拓扑序列，每次选中某个点后要将它所指向的所有结点的入度-1，最后根据是否出现过入度不为0的点决定是否要输出当前的编号i

3.flag是用来判断之前是否输出过现在是否要输出空格的

4.judge是用来判断是否是拓扑序列的

# 满分代码 #

#include <iostream>
    #include <vector>
    using namespace std;
    int main() {
        int n, m, a, b, k, flag = 0, in[1010];
        vector<int> v[1010];
        scanf("%d %d", &n, &m);
        for (int i = 0; i < m; i++) {
            scanf("%d %d", &a, &b);
            v[a].push_back(b);
            in[b]++;
        }
        scanf("%d", &k);
        for (int i = 0; i < k; i++) {
            int judge = 1;
            vector<int> tin(in, in+n+1);
            for (int j = 0; j < n; j++) {
                scanf("%d", &a);
                if (tin[a] != 0) judge = 0;
                for (int it : v[a]) tin[it]--;
            }
            if (judge == 1) continue;
            printf("%s%d", flag == 1 ? " ": "", i);
            flag = 1;
        }
        return 0;
    }

特别推荐：[任务调度的合理性 (25分)][_25]

备考汇总贴：[2020年6月PAT甲级满分必备刷题技巧][2020_6_PAT]

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[https_pintia.cn_problem-sets_994805342720868352_problems_994805343043829760]: https://pintia.cn/problem-sets/994805342720868352/problems/994805343043829760
[2020_6_PAT]: https://blog.csdn.net/allisonshing/article/details/104213959
[gre.jpg]: /images/20230529/8ec25c408a0c45ff90d452d06dfb71b7.png
[liuchuo]: https://www.liuchuo.net/archives/4665
[_25]: https://blog.csdn.net/allisonshing/article/details/104910631