poj3249 拓扑找最长路

Test for Job

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 11230		Accepted: 2651

Description

Mr.Dog was fired by his company. In order to support his family, he must find a new job as soon as possible. Nowadays, It's hard to have a job, since there are swelling numbers of the unemployed. So some companies often use hard tests for their recruitment.

The test is like this: starting from a source-city, you may pass through some directed roads to reach another city. Each time you reach a city, you can earn some profit or pay some fee, Let this process continue until you reach a target-city. The boss will compute the expense you spent for your trip and the profit you have just obtained. Finally, he will decide whether you can be hired.

In order to get the job, Mr.Dog managed to obtain the knowledge of the net profit V_i of all cities he may reach (a negative V_i indicates that money is spent rather than gained) and the connection between cities. A city with no roads leading to it is a source-city and a city with no roads leading to other cities is a target-city. The mission of Mr.Dog is to start from a source-city and choose a route leading to a target-city through which he can get the maximum profit.

Input

The input file includes several test cases.
The first line of each test case contains 2 integers n and m(1 ≤ n ≤ 100000, 0 ≤ m ≤ 1000000) indicating the number of cities and roads.
The next n lines each contain a single integer. The ith line describes the net profit of the city i, V_i (0 ≤ |V_i| ≤ 20000)
The next m lines each contain two integers x, y indicating that there is a road leads from city x to city y. It is guaranteed that each road appears exactly once, and there is no way to return to a previous city.

Output

The output file contains one line for each test cases, in which contains an integer indicating the maximum profit Dog is able to obtain (or the minimum expenditure to spend)

Sample Input

6 5
1
2
2
3
3
4
1 2
1 3
2 4
3 4
5 6

Sample Output

7

Hint

 

注意是无环图从入度为0出发至出度为0

无环图知每个点只能用一次，然后用的时候显然是有顺序的，那么就可以根据入度来排序从而推进答案喽

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + ;
int n, m, tot;
int cost[N], in[N], out[N], head[N], dp[N];
bool vis[N];
struct node
{
 int to, nxt;
}e[N * ];
void add(int x, int y)
{
 e[tot].to = y;
 e[tot].nxt = head[x];
 head[x] = tot++;
}
void toposort()
{
 int cnt = ;
 while(cnt < n) {
  for(int i = ; i <= n; ++i)
   if(in[i] ==  && !vis[i]) {
    vis[i] = true;
    cnt++;
    for(int j = head[i]; j != -; j = e[j].nxt) {
     int x = e[j].to;
     in[x]--;
     if(dp[i] + cost[x] > dp[x]) dp[x] = dp[i] + cost[x];
    }
   }
 }
}
int main()
{
 while(scanf("%d%d", &n, &m) != EOF) {
  memset(in, , sizeof(in));
  memset(out, , sizeof(out));
  memset(head, -, sizeof(head));
  memset(vis, false, sizeof(vis));
  tot = ;
  for(int i = ; i <= n; ++i)
   scanf("%d", &cost[i]);
  for(int i = ; i <= m; ++i) {
   int x, y;
   scanf("%d%d", &x, &y);
   add(x, y);
   in[y]++;
   out[x]++;
  }
  for(int i = ; i <= n; ++i)
   if(in[i] == ) dp[i] = cost[i];
   else dp[i] = -INF;
  toposort();
  int ans = -INF;
  for(int i = ; i <= n; ++i) if(out[i] ==  && dp[i] > ans) ans = dp[i];
  printf("%d\n", ans);
 }
 return ;
}