POJ-2387-Til the Cows Come Home（最短路）

缺乏、安全感 2021-09-20 12:40 220阅读 0赞

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.  
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

--------------------

思路：点对点最短路问题，dijkstra算法就OK；

坑点：

1.两棵树之间可能有多条路径，输入时取最小的。

2.路均是双向路，所以输出时需  G\[v\]\[u\]=G\[u\]\[v\]=w;

--------------------

1 #include<cstdio>
     2 #include<cstring>
     3 using namespace std;
     4 const int Inf=0x3f3f3f3f,N=1e3+5;
     5 
     6 int G[N][N],dis[N],mark[N];
     7 int n,m;
     8     
     9 void Getmap(){
    10     memset(G,Inf,sizeof(G));
    11     for(int i=0;i<=n;i++)
    12         G[i][i]=0;
    13         
    14     for(int i=0;i<m;i++){
    15         int u,v,w;
    16         scanf("%d%d%d",&u,&v,&w);
    17         if(G[u][v]>w)//两城之间可能有多条路，输入最短的 
    18             G[v][u]=G[u][v]=w;//双向路 
    19     }
    20 }
    21 
    22 void Dijk(){
    23     memset(mark,0,sizeof(mark));
    24     for(int i=1;i<=n;i++)
    25         dis[i]=G[1][i];
    26     mark[1]=1;
    27     for(int k=1;k<=n;k++){
    28          int mini=Inf;
    29          int u;
    30         for(int i=1;i<=n;i++)
    31             if(!mark[i]&&dis[i]<mini){
    32                 mini=dis[i];
    33                 u=i;
    34             }
    35         mark[u]=1;
    36         for(int i=1;i<=n;i++)
    37             if(dis[i]>dis[u]+G[u][i])
    38                 dis[i]=dis[u]+G[u][i];
    39     }     
    40 }
    41 
    42 int main(){
    43     while(~scanf("%d%d",&m,&n)){
    44         
    45     Getmap();
    46     Dijk();
    47     printf("%d\n",dis[n]);
    48     
    49     }
    50     return 0;
    51 }

转载于:https://www.cnblogs.com/yzhhh/p/9977746.html