POJ 3249 Test for Job (记忆化搜索 好题)

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.

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
int const MAX = 1e5 + 5;
int const INF = 2147483640;
int n, m;
int dp[MAX], val[MAX], ind[MAX];
vector <int> vt[MAX];

int DFS(int x)
{
    if(dp[x])
        return dp[x];
    int sz = vt[x].size();
    if(sz == 0)
        return val[x];
    int tmp = -INF;
    for(int i = 0; i < sz; i++)
    {
        int u = vt[x][i];
        tmp = max(tmp, val[x] + DFS(u));
    }
    return dp[x] = tmp;
}

int main()
{
    while(scanf("%d %d", &n, &m) != EOF)
    {
        for(int i = 1; i <= n; i++)
            vt[i].clear();
        memset(ind, 0, sizeof(ind));
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++)
            scanf("%d", &val[i]);
        for(int i = 0; i < m; i++)
        {
            int u, v;
            scanf("%d %d", &u, &v);
            vt[u].push_back(v);
            ind[v]++;
        }
        int ans = -INF;
        for(int i = 1; i <= n; i++)
            if(!ind[i])
                ans = max(ans, DFS(i));
        printf("%d\n", ans);
    }
}