POJ3249 Test for Job DAG最短路

　　题目链接：http://poj.org/problem?id=3249

　　DAG图上的最短路，记忆化搜索。

//STATUS:C++_AC_2000MS_14272KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=100010;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

struct Edge{
    int u,v;
}e[N*10];
int first[N],next[N*10],val[N],cnti[N],cnto[N];
LL d[N];
int n,m,mt;

void adde(int a,int b)
{
    e[mt].u=a;e[mt].v=b;
    next[mt]=first[a];first[a]=mt++;
}

LL dfs(int u)
{
    if(d[u]!=-LNF)return d[u];
    if(!cnto[u])return d[u]=val[u];
    int i;
    for(i=first[u];i!=-1;i=next[i]){
        d[u]=Max(d[u],(LL)val[u]+dfs(e[i].v));
    }
    return d[u];
}

int main()
{
  //  freopen("in.txt","r",stdin);
    int i,j,a,b;
    LL ans;
    while(~scanf("%d%d",&n,&m))
    {
        for(i=1;i<=n;i++)
            scanf("%d",&val[i]);
        mt=0;
        mem(first,-1);
        mem(cnti,0);mem(cnto,0);
        for(i=1;i<=m;i++){
            scanf("%d%d",&a,&b);
            cnti[b]++,cnto[a]++;
            adde(a,b);
        }

        ans=-LNF;
        for(i=1;i<=n;i++)d[i]=-LNF;
        for(i=1;i<=n;i++){
            if(!cnti[i]){
                ans=Max(ans,dfs(i));
            }
        }

        printf("%I64d\n",ans);
    }
    return 0;
}