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POJ 1860 SPFA 最长路

货币兑换问题,经典问题,这个问题解决的关键是发现,如果存在正环,那么一定是YES。稍微改一下SPFA,寻找一个图中的正环。

 

   1:  /**
   2:  search the longest path , just jude whether there are a positve cycle.
   3:  
   4:  */
   5:   
   6:  #include <queue>
   7:  #include <iostream>
   8:  #include <string.h>
   9:  #include <stdio.h>
  10:  using namespace std;
  11:  #define V 3000      // vertex
  12:  #define E 1500      // edge
  13:  #define INF 1000000.0f
  14:   
  15:  struct node
  16:  {
  17:      int v, next;
  18:      double R,C,w;
  19:  }pnt[E];
  20:   
  21:  int head[V];
  22:  double  dis[V];
  23:  bool vis[V];
  24:  int  cnt[V];          // the num of the operation of push to Quque. negatvie cycle.
  25:  int e = 0;            // the index of the edge
  26:  int N ;               // the num of the vertex.
  27:  int M ;               // the num of edges
  28:  int src, sink;
  29:  double val;
  30:  void addedge(int  u, int v, double R, double C)
  31:  {
  32:      pnt[e].v = v; pnt[e].w= 0;
  33:      pnt[e].R = R; pnt[e].C = C;
  34:      pnt[e].next = head[u]; head[u] = e++;
  35:  }
  36:  void SPFA_init()
  37:  {
  38:      e=0;
  39:      memset(head, -1,sizeof(head));
  40:      memset(vis, 0 ,sizeof(vis));
  41:      memset(cnt, 0 ,sizeof(cnt));
  42:      for(int i=0; i<=N; i++) dis[i] = 0;
  43:  }
  44:  int SPFA()
  45:  {
  46:      queue<int> Q;
  47:      Q.push(src); vis[src] = 1; dis[src] = val; ++cnt[src]; 
  48:      while(!Q.empty()) 
  49:      {
  50:          int u = Q.front(); Q.pop();  vis[u] = 0;
  51:          for(int i=head[u]; i!=-1; i=pnt[i].next)
  52:          {
  53:              int v = pnt[i].v;
  54:              pnt[i].w = (dis[u] - pnt[i].C)*pnt[i].R - dis[u];
  55:              //cout<<u<<" to "<<v<<" "<<pnt[i].w<<endl;
  56:              if( dis[v] < dis[u] + pnt[i].w  )
  57:              {
  58:                  dis[v] = dis[u] + pnt[i].w;
  59:                  //cout<<u<<" to "<<v<<"  "<<pnt[i].w<<" "<<dis[v]<<endl;
  60:                  if(!vis[v]) { Q.push(v); vis[v] = 1;}
  61:                  if( ++cnt[v] > N) return -1; // negative cycle.
  62:              }
  63:          }
  64:      }
  65:      return 1;
  66:      //if(dis[sink] == INF) return -2;          // can't from src to sink. 
  67:      //return dis[sink];
  68:  }
  69:   
  70:  int main()
  71:  {
  72:      freopen("1860.txt","r",stdin);
  73:      scanf("%d%d", &N , &M);
  74:   
  75:      scanf("%d", &src);
  76:      //cout<<src<<endl;
  77:      scanf("%lf", &val);
  78:      //cout<<val<<endl;
  79:      SPFA_init();
  80:      for(int i=0; i<M; i++)
  81:      {
  82:          int a, b; double Rab, Cab, Rba,Cba;
  83:          scanf("%d%d", &a, &b);
  84:          scanf("%lf%lf", &Rab, &Cab);
  85:          scanf("%lf%lf", &Rba, &Cba);
  86:          addedge(a, b,Rab, Cab);
  87:          addedge(b,a,Rba, Cba);
  88:          //cout<<Rab<<" "<<Cab<<" "<<Rba<<" "<<Cba<<endl;
  89:      }
  90:      int ret = SPFA();
  91:      if(ret == -1)  cout<<"YES"<<endl;
  92:      else cout<<"NO"<<endl;
  93:      return 0;
  94:  }

posted on 2012-11-09 23:24 Sosi 阅读(481) 评论(0)  编辑 收藏 引用 所属分类: Algorithm


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