HDU 5195 DZY Loves Topological Sorting 拓扑排序

 1 #include<algorithm>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<vector>
 6 #include<queue>
 7 using namespace std;
 8 
 9 const int maxn = 1e5 + 10;
10 const int INF = 0x3f3f3f3f;
11 
12 int n, m, k;
13 
14 struct Node {
15     int v, flag;
16     Node(int v,int flag=0):v(v),flag(flag){}
17     Node() { flag = 0; }
18 };
19 
20 vector<Node> head[maxn];
21 int ind[maxn],done[maxn];
22 
23 void init() {
24     for (int i = 0; i < n; i++) {
25         ind[i] = done[i]=0;
26         head[i].clear();
27     }
28 }
29 
30 int main() {
31     while (scanf("%d%d%d", &n, &m, &k) == 3 && n) {
32         init();
33         for (int i = 0; i < m; i++) {
34             int u, v;
35             scanf("%d%d", &u, &v); u--, v--;
36             ind[v]++;
37             head[u].push_back(Node(v,0));
38         }
39 
40         priority_queue<int> pq;
41 
42         //删边
43         for (int i = n - 1; i >= 0; i--) {
44             if (k >= ind[i]) {
45                 //将i的父亲ui中，满足ui<i，即边（ui,i）删了，这里要注意，对于边（ui,i),ui>i的边，根本不用删
46                 k -= ind[i];
47                 pq.push(i);
48                 for (int j = 0; j < head[i].size(); j++) {
49                     Node &nd = head[i][j];
50                     if (i > nd.v) {
51                         //把边（i,v）删了，拓扑排序的时候不能再走这条边了
52                         nd.flag = 1;
53                         ind[nd.v]--;
54                     }
55                 }
56             }
57         }
58 
59         vector<int> ans;
60         //拓扑排序
61         while (!pq.empty()) {
62             int u = pq.top(); pq.pop();
63             if (done[u]) continue;
64             ans.push_back(u);
65             done[u] = 1;
66             for (int i = 0; i < head[u].size(); i++) {
67                 Node& nd = head[u][i];
68                 if (done[nd.v]||nd.flag) continue;
69                 ind[nd.v]--;
70                 if (ind[nd.v] == 0) pq.push(nd.v);
71             }
72         }
73 
74         printf("%d", ans[0]+1);
75         for (int i = 1; i < ans.size(); i++) printf(" %d", ans[i]+1);
76         printf("\n");
77     }
78     return 0;
79 }
80 /*
81 5 3 1
82 4 3
83 1 3
84 3 2
85 
86 5 3 0
87 4 3
88 1 3
89 3 2
90 */