POJ 3249 Test for Job【SPFA】

Test for Job

Time Limit: 5000MS

 

Memory Limit: 65536K

Total Submissions: 10381

 

Accepted: 2421

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

Source

POJ Monthly--2007.07.08,落叶飞雪

 

 题目大意：给出n个点，m条边，每个点都提供了相对的点权值，然后给出相连着的边，问最大利润值。

 思路：

1、SPFA求最长路，正向建边TLE,反向建边4000msAC。。。。

2、这里源点一定是入度为0的点，因为题干中说了，没有其他任何点有边有有向边连接源点，那么起点的选取一定是要入度为0的点才行，相应的，终点也一定是出度为0的点才行。

 3、卡vector、卡cin、习惯用这两种来实现代码的，请使用链式前向星和scanf。

 

 AC代码：

#include<stdio.h>#include<string.h>#include<queue>using namespace std;int head[200005];struct EdgeNode{    int to;    int w;    int next;}e[200005*20];int degree[200005];int val[200005];int vis[200005];int dis[200005];int n,m;int cont;void add(int from,int to){    e[cont].to=to;    e[cont].next=head[from];    head[from]=cont++;}void SPFA(){    queue<int >s;    for(int i=0;i<n;i++)dis[i]=-0x3f3f3f3f;    for(int i=0;i<n;i++)    {        if(degree[i]==0)//处理源点。        {            dis[i]=val[i];            vis[i]=1;            s.push(i);        }    }    while(!s.empty())    {        int u=s.front();        s.pop();vis[u]=0;        for(int k=head[u];k!=-1;k=e[k].next)        {            int v=e[k].to;            if(dis[v]<dis[u]+val[v])//求最长路            {                dis[v]=dis[u]+val[v];                if(vis[v]==0)                {                    vis[v]=1;                    s.push(v);                }            }        }    }    int output=-0x3f3f3f3f;    for(int i=0;i<n;i++)    {        if(head[i]==-1)        output=max(output,dis[i]);    }    printf("%d\n",output);}int main(){    while(~scanf("%d%d",&n,&m))    {        cont=0;        memset(vis,0,sizeof(vis));        memset(degree,0,sizeof(degree));        memset(head,-1,sizeof(head));        for(int i=0;i<n;i++)        {            scanf("%d",&val[i]);        }        for(int i=0;i<m;i++)        {            int x,y;            scanf("%d%d",&x,&y);            x--;y--;            add(y,x);            degree[x]++;        }        SPFA();    }}