PKU2387-Til the Cows Come Home（SPFA+邻接表）

这道题还是蛮经典的图论题。

题意：

Description

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. 

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it. 

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.
       由bidirectional（双向）可知此图是无向图。
       要求你输入ne（边数）和nv（顶点数），然后是每条路径s到t的权值w.

       然后小女孩从最后一个节点回到起点，求最短路。

解题思路：

             点的个数为1000，不算很大，所以邻接矩阵跟邻接表都可以。下面我用邻接表。

代码：

 

#include<iostream>    using namespace std;    #define MAXE 10000  #define MAXV 1001  int inf = 1000000000;    struct edge_t  {      int s, t;      int w;      int next;  };    edge_t arrEdge[MAXE]; //arrEdge[1...szEdge]保存szEdge条边.  int szEdge;//主函数中方便记录边的位置  int arrHead[MAXV];//用来记录边的位置    int nv;  int arrDis[MAXV];  int Que[MAXE*10]; //队列  char inQue[MAXV]; //点v是否在队列中    void SPFA(int s)  {      int i, ie, v, head, tail;      edge_t e;        for(i = 1; i <= nv; i++)          arrDis[i] = inf;        arrDis[s] = 0;        //初始化队列      memset(Que, 0, sizeof(Que));      memset(inQue, 0, sizeof(inQue));      head = 0, tail = 1;      Que[head] = s;      inQue[s] = 1;      while(head < tail)      {          v = Que[head++];          inQue[v] = 0;            for(ie = arrHead[v]; ie!=-1; ie = arrEdge[ie].next)          {              e = arrEdge[ie];              if(arrDis[e.t] > arrDis[v] + e.w)//当这里不能在松弛时，不用入队，即算法结束              {                  arrDis[e.t] = arrDis[v] + e.w;                  if(!inQue[e.t])                  {                      inQue[e.t] = 1;                      Que[tail++] = e.t;                  }              }          }      }      cout<<arrDis[1]<<endl;      return ;  }    int main()  {      int ne, i, s, t, w, p;        memset(arrHead, -1, sizeof(arrHead));      memset(arrEdge, 0, sizeof(arrEdge));      szEdge = 0;        scanf("%d %d", &ne, &nv);      for(i = 1; i <= ne; i++)      {          scanf("%d %d %d", &s, &t, &w);          szEdge++;          arrEdge[szEdge].s = s;          arrEdge[szEdge].t = t;          arrEdge[szEdge].w = w;          arrEdge[szEdge].next = arrHead[s];          arrHead[s] = szEdge;            szEdge++;          arrEdge[szEdge].s = t;          arrEdge[szEdge].t = s;          arrEdge[szEdge].w = w;          arrEdge[szEdge].next = arrHead[t];          arrHead[t] = szEdge;      }        SPFA(nv);        return 0;  }