【HDOJ】4857 逃生

很容易想到优先队列+拓扑排序。关键点是有限制条件者有限，无限制条件者在最后，条件相同者按序输出。因此采用逆序。

 1 #include <iostream>
 2 #include <queue>
 3 #include <algorithm>
 4 #include <cstdio>
 5 #include <cstring>
 6 using namespace std;
 7
 8 #define MAXN 30005
 9
10 typedef struct {
11     int to, next;
12 } edge_st;
13
14 edge_st edges[100005];
15 int adj[MAXN], L;
16 int deg[MAXN];
17 int buf[MAXN];
18
19 typedef struct node_st{
20     int i;
21     node_st() {}
22     node_st(int ii) {
23         i = ii;
24     }
25     friend bool operator <(node_st a, node_st b) {
26         return a.i < b.i;
27     }
28 } node_st;
29
30 void add(int u, int v) {
31     edges[L].to = v;
32     edges[L].next = adj[u];
33     adj[u] = L++;
34     ++deg[v];
35 }
36
37 void init() {
38     L = 1;
39     memset(deg, 0, sizeof(deg));
40     memset(adj, 0, sizeof(adj));
41 }
42
43 int main() {
44     int T, n, m;
45     int i, j, k;
46     node_st nd;
47
48     scanf("%d", &T);
49     while (T--) {
50         scanf("%d%d", &n, &m);
51         init();
52         while (m--) {
53             scanf("%d %d", &j, &k);
54             add(k, j);
55         }
56         priority_queue<node_st> Q;
57         for (i=1; i<=n; ++i)
58             if (deg[i] == 0)
59                 Q.push(node_st(i));
60         k = 0;
61         while (!Q.empty()) {
62             nd = Q.top();
63             buf[k++] = nd.i;
64             Q.pop();
65             for (i=adj[nd.i]; i; i=edges[i].next) {
66                 deg[edges[i].to]--;
67                 if (deg[edges[i].to] == 0)
68                     Q.push(node_st(edges[i].to));
69             }
70         }
71         for (i=k-1; i>=0; --i)
72             if (i)
73                 printf("%d ", buf[i]);
74             else
75                 printf("%d\n", buf[i]);
76     }
77
78     return 0;
79 }