数据结构 课程安排 (拓扑排序)

Description

针对计算机系本科课程，根据课程之间的依赖关系（如离散数学应在数据结构之前开设）制定课程安排计划。

Input

第一行为样例组数T。每组样例第一行为课程数量n（1 <= n <= 5000)，以下n行每行表示一门课程名称。接下来为关系数量m（1 <= m <= 10000)，每一行有两个课程名称a、b，表示a课程要开设在b课程前面。（输入保证无环）

Output

每组样例第一行见输出，以下n行每行输出一个课程名称，描述拓扑排序后的课程表。如果课程优先级相同，则优先输出课程名称字典序小的课程。

Sample Input

1

6

Math

Chinese

English

Sports

Music

Computer

4

Chinese English

English Computer

Math Computer

Music Computer

Sample Output

Case 1:

Chinese English

Math Music

Computer Sports

HINT

考查知识点：拓扑排序

Append Code

析：拓扑排序，先把所有的科目都排序，按照字典序排，然后再编号，在拓扑排序时，编号小的优先，可用优先队列实现，其他的就很简单了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e3 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const int hr[]= {-2, -2, -1, -1, 1, 1, 2, 2};
const int hc[]= {-1, 1, -2, 2, -2, 2, -1, 1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
 
}
struct Node{
    int to, next;
};
Node a[10005<<1];
int head[maxn];
map<string, int> mp;
string s[maxn];
int cnt, in[maxn], ans[maxn];
 
void add(int u, int v){
    a[cnt].to = v;
    a[cnt].next = head[u];
    head[u] = cnt++;
}
 
int main(){
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d", &n);
        mp.clear();
        char str[20];
        for(int i = 0; i < n; ++i){
            scanf("%s", str);
            s[i] = str;
            head[i] = -1;
            in[i] = 0;
        }
        sort(s, s+n);
        for(int i = 0; i < n; ++i)  mp[s[i]] = i;
        scanf("%d", &m);
        cnt = 0;
        for(int i = 0; i < m; ++i){
            scanf("%s", str);
            int u = mp[str];
            scanf("%s", str);
            int v = mp[str];
            add(u, v);
            ++in[v];
        }
        priority_queue<int, vector<int>, greater<int> >pq;
        for(int i = 0; i < n; ++i) if(!in[i])  pq.push(i);
        int tot = 0;
        while(!pq.empty()){
            int x = pq.top();  pq.pop();
            ans[tot++] = x;
            for(int i = head[x]; ~i; i = a[i].next){
                int t = a[i].to;
                --in[t];
                if(!in[t]) pq.push(t);
            }
        }
        printf("Case %d:\n", kase);
        for(int i = 0; i < n; ++i)
            printf("%s\n", s[ans[i]].c_str());
 
    }
    return 0;
}