Til the Cows Come Home 【SPFA】 【djk】

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.
Input
* Line 1: Two integers: T and N

• Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.
  Output
• Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.
  Sample Input
  5 5
  1 2 20
  2 3 30
  3 4 20
  4 5 20
  1 5 100
  Sample Output
  90
  Hint
  INPUT DETAILS:

There are five landmarks.

OUTPUT DETAILS:

Bessie can get home by following trails 4, 3, 2, and 1.

不多说，模版题，只是为了熟悉下两种算法
第一种 SPFA + 邻接表 （建图快一些） 63 ms
代码

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>#include<queue>#include<stack>#include<map>#include<vector>#include<set>#define CLR(a,b) memset((a),(b),sizeof(a))#define inf 0x3f3f3f3f#define mod 100009#define LL long long#define M  10000#define ll o<<1#define rr o<<1|1#define lson o<<1,l,mid#define rson o<<1|1,mid+1,rusing namespace std;struct edge {int to,cost;};vector<edge >G[M];int dis[M];int n,m;void init(){ for(int i=1;i<=n;i++) dis[i]=inf,G[i].clear(); }void getmap(){    int i,j;    for(i=1;i<=m;i++)    {        int a,b,c;        scanf("%d%d%d",&a,&b,&c);        G[a].push_back({b,c});        G[b].push_back({a,c});    }}void spfa(int st,int ed){    dis[st]=0;int i,j;    queue<int>Q;    Q.push(st);    while(!Q.empty())    {        int next=Q.front();Q.pop();        for(i=0;i<G[next].size();i++)        {            edge e=G[next][i];            if(dis[e.to]>dis[next]+e.cost)            {                dis[e.to]=dis[next]+e.cost;                Q.push(e.to);            }        }     }     printf("%d\n",dis[ed]) ;}int main(){    scanf("%d%d",&m,&n);    init();    getmap();    spfa(n,1);    return 0;}

第二种 DJK + 邻接表 用优先队列维护的，时间复杂度为 E*logV 94ms
代码

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>#include<queue>#include<stack>#include<map>#include<vector>#include<set>#define CLR(a,b) memset((a),(b),sizeof(a))#define inf 0x3f3f3f3f#define mod 100009#define LL long long#define M  10000#define ll o<<1#define rr o<<1|1#define lson o<<1,l,mid#define rson o<<1|1,mid+1,rusing namespace std;typedef struct pair <int ,int >P;struct edge{int to,cost;};vector<edge>G[M];int dis[M];int n,m;void init(){ for(int i=1;i<=n;i++) G[i].clear(),dis[i]=inf;}void getmap(){    int i,j;    for(i=0;i<m;i++)    {        int a,b,c;        scanf("%d%d%d",&a,&b,&c);        G[a].push_back({b,c});        G[b].push_back({a,c});    }}void djk(int st,int ed){    int i,j;    priority_queue<P ,vector <  P > ,greater  < P  > > Q;    dis[st]=0;Q.push({0,st});    while(!Q.empty())    {        P now = Q.top();Q.pop();        int value=now.first;int next=now.second;        if(dis[next]<value) continue;        for(i=0;i<G[next].size();i++)        {            edge e=G[next][i];            if(dis[next]+e.cost<dis[e.to])            {                dis[e.to]=dis[next]+e.cost;                Q.push({dis[e.to],e.to});                }           }    }    printf("%d\n",dis[ed]);}int main(){    scanf("%d%d",&m,&n);    init();    getmap();    djk(n,1);    return 0;}

第三种 SPFA + 链式向前星 （手动建链表） 不到 20ms
代码

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>#include<queue>#include<stack>#include<map>#include<vector>#include<set>#define CLR(a,b) memset((a),(b),sizeof(a))#define inf 0x3f3f3f3f#define mod 100009#define LL long long#define M 100000#define ll o<<1#define rr o<<1|1#define lson o<<1,l,mid#define rson o<<1|1,mid+1,rusing namespace std;int n,m;struct edge{int to,cost,next;}G[M];int dis[M],vis[M],head[M],top;void addedge(int a,int b,int c){    G[top].to=b;    G[top].cost=c;    G[top].next=head[a];    head[a]=top++;}void init(){    top=0;    for(int i=1;i<=n;i++)  // 这里决定了，点的序号是从 0 or 1 开始的      {        head[i]=-1;        vis[i]=0;        dis[i]=inf;    }}void getmap(){    int i,j;    for(i=0;i<m;i++)    {        int a,b,c;        scanf("%d%d%d",&a,&b,&c);        addedge(a,b,c);        addedge (b,a,c);    }}void spfa(int st,int ed){    dis[st]=0;vis[st]=1;    queue<int>Q;    Q.push(st);    while(!Q.empty())    {        int now=Q.front();Q.pop();        vis[now]=0;        for(int i=head[now];i!=-1;i=G[i].next)        {            int next=G[i].to;            if(dis[next]>dis[now]+G[i].cost)            {                dis[next]=dis[now]+G[i].cost;                if(!vis[next])  //  小优化，不会让多余的点进                {                    vis[next]=1;                    Q.push(next);                }            }        }    }    printf("%d\n",dis[ed]);}int main(){    scanf("%d%d",&m,&n);    init();    getmap();    spfa(n,1);    return 0;}

第四种 djk + 邻接矩阵 当图的边比较稠密时候，这个会更快 //93ms
代码

#include<stdio.h>  #include<string.h>  #include<algorithm>  #include<math.h>  #define inf 0x3f3f3f  #define mod 1000007  #define LL long long   #define M 1000+10   using namespace std;  int t,n;  int map[M][M];  int v[M];  int dis[M];  void getmap()  {      int i,j;      for(i=1;i<=n;i++)      for(j=1;j<=n;j++)      if(j==i) map[i][j]=0;      else map[i][j]=map[j][i]=inf;      for(i=1;i<=t;i++)      {          int a,b,c;          scanf("%d%d%d",&a,&b,&c);          if(map[a][b]>c)          map[a][b]=map[b][a]=c;      }  }    void djk(int st,int ed)  {      int i,j,k,u;      int min,next;      for(i=1;i<=n;i++)      {          v[i]=0;          dis[i]=map[st][i];       }       v[st]=1;       for(u=2;u<=n;u++)       {          min=inf;          for(j=1;j<=n;j++)          {              if(!v[j]&&dis[j]<min)              {                  min=dis[j];                  next=j;               }          }               if(min==inf)                break;                v[next]=1;                for(j=1;j<=n;j++)                {                  if(!v[j]&&dis[j]>dis[next]+map[next][j])                  dis[j]=dis[next]+map[next][j];                }       }       printf("%d\n",dis[ed]);  }  int main()  {      scanf("%d%d",&t,&n);      getmap();      djk(n,1);      return 0;   }