POJ 3169 差分约束方程

POJ 3169 差分约束方程

一道简单的差分约束方程，差分约束方程由Bellmanford转换而来，一组差分方程由 x-y <=d 的形式，而在最短路中松弛条件有dis(x)<= dis(y)+w。 所以产生了一条由y连向x的有向边，权重为d。

https://github.com/Sosi/ProgrammingContest/blob/master/OnlineJudge/POJ/PKU3169.cpp

   1:   

   2:  #include <queue>

   3:  #include <iostream>

   4:  #include <string.h>

   5:  #include <stdio.h>

   6:  using namespace std;

   7:  #define MAXN 1005      // vertex

   8:  #define MAXM 20005      // edge

   9:  #define INF 0x3F3F3F3F

  10:   

  11:  struct node

  12:  {

  13:      int v, w, next;

  14:  }pnt[MAXM];

  15:   

  16:  int head[MAXN];

  17:  int  dis[MAXN];

  18:  bool vis[MAXN];

  19:  int  cnt[MAXN];       // the number of the operation of push to Quque. negatvie cycle.

  20:  int num = 0;          // the index of the edge

  21:  int N ;               // the number of the vertex.

  22:  int M ;               // the number of edges

  23:  int src, sink;

  24:  void addedge(int  u, int v, int w)

  25:  {

  26:      pnt[num].v = v; pnt[num].w= w;

  27:      pnt[num].next = head[u]; head[u] = num++;

  28:  }

  29:   

  30:  int SPFA()

  31:  {

  32:      for(int i=0; i<=N; i++)

  33:      {

  34:          vis[i]=0; dis[i] = INF; cnt[i] = 0;

  35:      }

  36:   

  37:      queue<int> Q;

  38:      Q.push(src); vis[src] = 1; dis[src] = 0; ++cnt[src]; 

  39:      while(!Q.empty()) 

  40:      {

  41:          int u = Q.front(); Q.pop();  vis[u] = 0;

  42:          for(int i=head[u]; i!=-1; i=pnt[i].next)

  43:          {

  44:              int v = pnt[i].v;

  45:              if( dis[v] > dis[u] + pnt[i].w  )

  46:              {

  47:                  dis[v] = dis[u] + pnt[i].w;

  48:                  if(!vis[v]) {Q.push(v); vis[v] = 1;}

  49:                  if( ++cnt[v] > N) return -1; // negative cycle.

  50:              }

  51:          }

  52:      }

  53:      if(dis[sink] == INF) return -2; // can't from src to sink.

  54:      return dis[sink];

  55:  }

  56:   

  57:  int main()

  58:  {

  59:      int ml,md;

  60:      while(scanf("%d%d%d", &N , &ml, &md)!= EOF)

  61:      {

  62:          num = 0;

  63:          memset(head, -1, sizeof(head)); 

  64:          int a, b, c;

  65:          for(int i=0; i<ml; i++)

  66:          {

  67:              scanf("%d%d%d", &a, &b, &c);

  68:              if(a>b) swap(a,b);

  69:              addedge(a, b,c);

  70:          }

  71:          for(int i=0; i<md; i++)

  72:          {

  73:              scanf("%d%d%d", &a, &b, &c);

  74:              if(a<b) swap(a,b);

  75:              addedge(a, b, -c);

  76:          }

  77:          src = 1; sink = N;

  78:          printf("%d\n", SPFA());

  79:      }

  80:      return 0;

  81:  }

.csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; }