PAT 1146. Topological Order (25) – 甲级

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.

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

题目大意：给一个有向图，判断给定序列是否是拓扑序列～
分析：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～flag是用来判断之前是否输出过现在是否要输出空格的～judge是用来判断是否是拓扑序列的～

#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;
}

#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;

}

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