POJ 3249 Test for Job DAG图单源最短路

Test for Job

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 10651		Accepted: 2504

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

明显 图是有向无环图
用拓扑排序一遍,然后按拓扑顺序对点的出边进行松弛操作,就能得到最短路
此题可以有多个起点 在构图的时候记录入度出度 入度为0的就是起点 出度为0的是终点
要注意的是最后ans只取出度为0的,dist初值要初始化为-INF,因为每个点的权值可正可负
拓扑排序用bfs会超时

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<math.h>
#include<list>
#include<cstring>
#include<fstream>
//#include<memory.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define INF 1000000007
#define pll pair<ll,ll>
#define pid pair<int,double>

char readT;
bool readIsNegative;

inline bool read(int&x){
    readIsNegative=x=0;
    while(!isdigit(readT=getchar()))
        if(readT==EOF)
            return false;
        else
            if(readT=='-'){
                readIsNegative=true;
                readT='0';
                break;
            }
    x=readT-'0';
    while(isdigit(readT=getchar()))
        x=x*10+readT-'0';
    if(readIsNegative)
        x=-x;
    return true;
}

const int N=100000+5;
const int M=1000000+5;

int head[N];
struct Edge{
    int to,next;
}edge[M];

inline void addEdge(int k,int u,int v){
    edge[k].to=v;
    edge[k].next=head[u];
    head[u]=k;
}

int w[N];
int dist[N];
int inDegree[N];
int outDegree[N];
bool used[N];
int TPO[N];
int TPOindex;

/*BFS TLE
void TopologicalOrder(int n,int index){
    if(n==index)
        return;
    for(int i=1;i<=n;++i){
        if(!used[i]&&inDegree[i]==0){
            for(int j=head[i];j!=-1;j=edge[j].next){
                --inDegree[edge[j].to];
            }
            used[i]=true;
            TPO[index]=i;
            TopologicalOrder(n,index+1);
            return;
        }
    }
}
*/

void TopologicalOrder(int u){
    used[u]=true;
    for(int i=head[u];i!=-1;i=edge[i].next){
        if(used[edge[i].to]==false)
            TopologicalOrder(edge[i].to);
    }
    TPO[TPOindex++]=u;
}

void relax(int n){
    TPOindex=0;
    for(int i=1;i<=n;++i){
        dist[i]=inDegree[i]==0?w[i]:-INF;
        if(used[i]==false)
            TopologicalOrder(i);
    }
    for(int i=n-1;i>=0;--i){
        int u=TPO[i];
        for(int j=head[u];j!=-1;j=edge[j].next){
            int v=edge[j].to;
            dist[v]=max(dist[u]+w[v],dist[v]);
        }
    }
}

int main()
{
    //freopen("/home/lu/文档/r.txt","r",stdin);
    //freopen("/home/lu/文档/w.txt","w",stdout);
    int n,m;
    //while(~scanf("%d%d",&n,&m)){
    while(read(n)){
        read(m);
        for(int i=1;i<=n;++i){
            read(w[i]);
            //scanf("%d",w+i);
            head[i]=-1;
            outDegree[i]=inDegree[i]=0;
            used[i]=false;
        }
        for(int i=0,u,v;i<m;++i){
            //scanf("%d%d",&u,&v);
            read(u);
            read(v);
            addEdge(i,u,v);
            ++inDegree[v];
            ++outDegree[u];
        }
        relax(n);
        int ans=-INF;
        for(int i=1;i<=n;++i){
            if(outDegree[i]==0)
                ans=max(ans,dist[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}