Poj 3249 Test for Job (DAG longest-path memory-based search solution)

Test for Job Time Limit:5000 MS Memory Limit:65536 K Total Submissions:8990 Accepted:2004 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 traffic, 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 profitViOf all cities he may reach (a negativeViIndicates 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 provided des several test cases. The first line of each test case contains 2 integersNAndM(1 ≤N≤ 100000, 0 ≤M≤ 1000000) indicating the number of cities and roads. The nextNLines each contain a single integer.ITh line describes the net profit of the cityI,Vi(0 ≤ |Vi| ≤ 20000) The next m lines each contain two integersX,YIndicating that there is a road leads from cityXTo cityY. 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 51223341 21 32 43 45 6 Sample Output 7 Hint Source POJ Monthly -- 2007.07.08, fallen leaf snow

Question:

To give you a graph, find a path from the starting point (the inbound degree is 0) to the ending point (the outbound degree is 0), and the sum of val of all vertices is the largest.

Ideas:

The first thought was SPFA, followed by endless TLE, and then searched by memory.

A simple small processing method is to add a Source Vertex to create an edge for all the starting points.

Code:

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #define maxn 100005#define MAXN 2000005#define INF 0x3f3f3f3ftypedef long long ll;using namespace std;ll n,m,ans,cnt,sx,oo;bool vis[maxn];ll dp[maxn],head[maxn];ll in[maxn],val[maxn];struct Node{    ll v,w,next;}edge[MAXN];void addedge(ll u,ll v,ll w){    cnt++;    edge[cnt].v=v;    edge[cnt].w=w;    edge[cnt].next=head[u];    head[u]=cnt;}ll dfs(ll u){    if(dp[u]!=oo) return dp[u];    ll i,j,t,v,best=oo,flg=0;    for(i=head[u];i;i=edge[i].next)    {        flg=1;        v=edge[i].v;        dfs(v);        best=max(best,dp[v]);    }    if(flg) dp[u]=best+val[u];    else dp[u]=val[u];}int main(){    ll i,j,u,v,w;    oo=-(1LL<<50);    while(~scanf("%lld%lld",&n,&m))    {        for(i=1;i<=n;i++)        {            scanf("%lld",&val[i]);        }        cnt=0;        for(i=0;i<=n;i++) head[i]=in[i]=0;        for(i=1;i<=m;i++)        {            scanf("%lld%lld",&u,&v);            addedge(u,v,val[u]);            in[v]++;        }        for(i=1;i<=n;i++)        {            if(in[i]==0) addedge(0,i,0);        }        for(i=0;i<=n;i++) dp[i]=oo;        dfs(0);        printf("%lld\n",dp[0]);    }    return 0;}