POJ 3249 Test for Job (dfs + dp)

题目链接：http://poj.org/problem?id=3249

题意:

给你一个DAG图，问你入度为0的点到出度为0的点的最长路是多少

思路：

记忆化搜索，注意v[i]可以是负的，所以初始值要-inf。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 const int N = 1e5 + 5;
 6 typedef long long LL;
 7 struct Edge {
 8     int next, to;
 9 }edge[N * 10];
10 LL dp[N], a[N], inf = 1e12;
11 int head[N], tot, input[N];
12
13 void init(int n) {
14     for(int i = 1; i <= n; ++i) {
15         head[i] = -1;
16         dp[i] = -inf;
17         input[i] = 0;
18     }
19     tot = 0;
20 }
21
22 inline void add(int u, int v) {
23     edge[tot].next = head[u];
24     edge[tot].to = v;
25     head[u] = tot++;
26 }
27
28 void dfs(int u) {
29     if(dp[u] != -inf)
30         return ;
31     for(int i = head[u]; ~i; i = edge[i].next) {
32         int v = edge[i].to;
33         dfs(v);
34         dp[u] = max(dp[v] + a[u], dp[u]);
35     }
36     if(dp[u] == -inf) //若是出度为0
37         dp[u] = a[u];
38 }
39
40 int main()
41 {
42     int n, m;
43     while(~scanf("%d %d", &n, &m)) {
44         for(int i = 1; i <= n; ++i) {
45             scanf("%lld", a + i);
46         }
47         init(n);
48         int u, v;
49         for(int i = 1; i <= m; ++i) {
50             scanf("%d %d", &u, &v);
51             add(u, v);
52             input[v]++;
53         }
54         LL Max = -inf;
55         for(int i = 1; i <= n; ++i) {
56             if(!input[i]) {
57                 dfs(i);
58                 Max = max(Max, dp[i]);
59             }
60         }
61         printf("%lld\n", Max);
62     }
63     return 0;
64 }