ZOJ 2770 Burn the Linked Camp

ZOJ 2770 Burn the Linked Camp

差分约束系统。
以下是我的代码：

/*
 * Author:  lee1r
 * Created Time:  2011/8/2 19:46:45
 * File Name: zoj2770.cpp
  */
#include < iostream >
#include < sstream >
#include < fstream >
#include < vector >
#include < list >
#include < deque >
#include < queue >
#include < stack >
#include < map >
#include < set >
#include < bitset >
#include < algorithm >
#include < cstdio >
#include < cstdlib >
#include < cstring >
#include < cctype >
#include < cmath >
#include < ctime >
#define  L(x) ((x)<<1)
#define  R(x) (((x)<<1)+1)
#define  Half(x) ((x)>>1)
#define  lowbit(x) ((x)&(-(x)))
using   namespace  std;
const   long   long  kInf(0x7f7f7f7f7f7f7f7fll);
const   double  kEps(1e - 11 );
typedef  long   long  int64;
typedef unsigned  long   long  uint64;

const   int  kMaxn( 1007 );
const   int  kMaxm( 20007 );

struct  Edge
{
     int  u,v,w;
};

int  n,m,cnt;
Edge e[kMaxm];
int64 ans;

void  AddEdge( int  u, int  v, int  w)
{
    cnt ++ ;
    e[cnt].u = u;
    e[cnt].v = v;
    e[cnt].w = w;
}

int64 BellmanFord()
{
    int64 d[kMaxn];
    d[n] = 0 ;
     for ( int  i = 0 ;i < n;i ++ )
        d[i] = kInf;
     for ( int  i = 0 ;i < n;i ++ )
         for ( int  j = 1 ;j <= cnt;j ++ )
             if (d[e[j].u] < kInf  &&  d[e[j].v] > d[e[j].u] + e[j].w)
                d[e[j].v] = d[e[j].u] + e[j].w;
     for ( int  i = 1 ;i <= cnt;i ++ )
         if (d[e[i].v] > d[e[i].u] + e[i].w)
             return   - 1 ;
     return  d[ 0 ];
}

int  main()
{   
     while (scanf( " %d%d " , & n, & m) == 2 )
    {
        cnt = 0 ;
         for ( int  i = 1 ;i <= n;i ++ )
        {
             int  t;
            scanf( " %d " , & t);
            AddEdge(i - 1 ,i,t);
            AddEdge(i,i - 1 , 0 );
        }
         for ( int  i = 1 ;i <= m;i ++ )
        {
             int  u,v,w;
            scanf( " %d%d%d " , & u, & v, & w);
            AddEdge(v,u - 1 , - w);
        }
        
        ans = BellmanFord();
        
         if (ans ==- 1 )
            printf( " Bad Estimations\n " );
         else
            cout <<- ans << endl;
    }
    
     return   0 ;
}