POJ 3249 （DAG）

Test for Job
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 11150 Accepted: 2630
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 Vi of all cities he may reach (a negative Vi 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, Vi (0 ≤ |Vi| ≤ 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

题意：给N个点，每个点的价值为Vi，有M条有向边，出度为0的是终点，入度为0的是起点，问从起点到终点经过点值和的最大值；

做法：记忆化搜索每个源点就OK了（开始没注意到V的取值还有负值就把ans初始化成了0结果wa了。。。）

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#define maxn 100010using namespace std;typedef long long ll;const ll inf = 2000000010;int head[maxn];int in[maxn];int out[maxn];ll v[maxn];ll dp[maxn];int cnt=0;int n,m;struct Node{    int fr,to,next;}e[1000010];void add(int a,int b){    e[cnt].fr=a;    e[cnt].to=b;    e[cnt].next=head[a];    head[a]=cnt++;}void init(){    memset(dp,0,sizeof(dp));    cnt=0;    memset(in,0,sizeof(in));    memset(out,0,sizeof(out));    memset(head,-1,sizeof(head));}int pre[maxn];ll dag(int x){    if(out[x]==0)return v[x];    if(dp[x]!=0)return dp[x];    ll ans=-inf;    for(int i=head[x];i!=-1;i=e[i].next)    {        int to=e[i].to;        pre[to]=x;        ll a=dag(to);        if(a+v[x]>ans)ans=a+v[x];    }    return dp[x]=ans;}int main(){    int a,b;    while(scanf("%d%d",&n,&m)==2)    {        init();        for(int i=1;i<=n;i++)        {            scanf("%I64d",&v[i]);        }        for(int i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            add(a,b);            out[a]++;            in[b]++;        }        ll ans=-inf;        for(int i=1;i<=n;i++)        {            if(in[i]==0)            {                memset(dp,0,sizeof(dp));                ll a=dag(i);                if(a>ans)ans=a;            }        }        printf("%I64d\n",ans);    }    return 0;}