POJ 3249 Test for Job (dfs + dp)
2016-10-11 17:03
423 查看
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。
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 , a , inf = 1e12; 11 int head , tot, input ; 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 }
相关文章推荐
- poj--3249 Test for Job(topsort + dp)
- poj 3249 Test for Job 图上dp(记忆化搜索)
- POJ 3249 Test for Job 拓扑图DP
- POJ 3249-Test for Job(拓扑排序&&DP)
- poj&nbsp;3249&nbsp;Test&nbsp;for&nbsp;Job&nbsp;dp(动态规…
- POJ 3249 Test for Job 解题报告 DP
- POJ 3249 Test For Job
- Test for Job (poj 3249 记忆化搜索)
- POJ 3249 Test for Job(记忆化搜索)
- POJ 3249 Test for Job【SPFA】
- poj 3249 Test for Job 最长路
- POJ 3249 Test for Job(记忆化搜索)
- POJ 3249 Test for Job(拓扑排序+dp)
- POJ 3249 Test for Job(拓扑排序)
- POJ - 3249 Test for Job (DAG+topsort)
- POJ-3249 Test for Job DAG最短路
- poj3249 Test for Job --- 拓扑排序
- poj 3249 Test for Job
- POJ 3249 Test for Job
- HDU 3249 Test for job (有向无环图上的最长路,DP)