POJ 3249 Test for Job 拓扑排序+DP

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<queue>
 5 using namespace std;
 6 
 7 int n, m, esize, x, y;
 8 int a[100100], en[100100], in[100100], bfn[100100], dp[100100];
 9 bool ed[100100];
10 struct edge{
11     int v, n;
12 } e[1000100];
13 void addedge(int u, int v)
14 {
15     e[esize].v = v;;
16     e[esize].n = en[u];
17     en[u] = esize ++;
18 }
19 queue<int> q;
20 int main()
21 {
22     while(scanf("%d%d", &n, &m) != EOF)
23     {
24         for (int i = 1; i <= n; i++)
25             scanf("%d", a+i);
26         memset(en, -1, sizeof(en));
27         memset(in, 0, sizeof(0));
28         memset(ed, true, sizeof(ed));
29         esize = 0;
30         for (int i = 0; i < m; i++){
31             scanf("%d %d", &x, &y);
32             addedge(x, y);
33             in[y]++;
34             ed[x] = 0;
35         }
36         for (int i = 1; i <= n; i++)
37             if (!in[i]) q.push(i);
38         int ans = -2147483640;
39         int cnt = 0;
40         while(!q.empty())
41         {
42             int u = q.front(); q.pop();
43             bfn[cnt++] = u;
44             for (int t = en[u]; t != -1; t = e[t].n){
45                 int v = e[t].v;
46                 in[v] --;
47                 if (in[v] == 0)    q.push(v);
48             }
49         }
50         memset(dp, -1, sizeof(dp));
51         for (int i = 0; i < n; i++){
52             int u = bfn[i];
53             if (dp[u] == -1) dp[u] = a[u];
54             if (ed[u]) ans = max(ans, dp[u]);
55             for (int t = en[u]; t != -1; t = e[t].n){
56                 int v = e[t].v;
57                 if (dp[v] == -1 || dp[v] < dp[u] + a[v])
58                     dp[v] = dp[u] + a[v];
59             }
60         }
61         printf("%d\n", ans);
62     }
63     return 0;
64 }