POJ-2387 Til the Cows Come Home（Bellman-Ford最短路）

忘是亡心i 2022-04-24 02:54 138阅读 0赞

**Problem 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:**

**Hint:**

INPUT DETAILS:   
  
There are five landmarks.   
  
OUTPUT DETAILS:   
  
Bessie can get home by following trails 4, 3, 2, and 1.

**思路：**

直接使用**Bellman-Ford算法**，将从1出发到达其他顶点的最短距离都计算出来保存在d数组中，最后输出下标为N的d数组元素的值（从编号为1的顶点出发到达编号为N的顶点的最短距离）。

**上AC代码：**

#include <stdio.h>
    typedef struct
    {
        int from;
        int to;
        int cost;
    }edge;
    const int INF=0x3f3f3f3f;
    edge bian[4002];
    int d[1001];
    int t,n;
    void shortestPath(int s)
    {
        int i;
        for(i=1;i<=n;i++)
        {
            d[i]=INF;
        }
        d[s]=0;
        while(true)
        {
            bool update=false;
            for(i=0;i<2*t;i++)
            {
                edge now=bian[i];
                if(d[now.from]!=INF&&d[now.to]>d[now.from]+now.cost)
                {
                    d[now.to]=d[now.from]+now.cost;
                    update=true;
                }
            }
            if(!update)
            {
                break;
            }
        }
    }
    int main()
    {
        scanf("%d%d",&t,&n);
        int i;
        for(i=0;i<t;i++)
        {
            scanf("%d%d%d",&bian[i].from,&bian[i].to,&bian[i].cost);
            bian[t+i].from=bian[i].to;
            bian[t+i].to=bian[i].from;
            bian[t+i].cost=bian[i].cost;
        }
        shortestPath(1);
        printf("%d\n",d[n]);
        return 0;
    }