A - Til the Cows Come Home (Dijkstra)

最新推荐文章于 2021-02-15 19:37:46 发布

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本文介绍了一种使用迪杰斯特拉算法求解最短路径的问题背景及实现过程。在一个包含多个地标点的农场中，Bessie需要找到从当前位置回到谷仓的最短路线。通过建立地标间的连接并赋予权重，运用迪杰斯特拉算法求得最优路径。

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.

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int ma[1281][2181];
int v[1281];
int dist[12123];
int m, n;
void Dijkstra(int x)
{
    memset(v, 0, sizeof(v));
    for(int i=1; i<=n; i++)
    {
        dist[i] = ma[x][i];
    }
    v[x] = 1;
    dist[x]= 0;
    for(int i=1; i<=n; i++)
    {
        int point = x;
        int min = 0x3f3f3f3f;
        for(int j=1; j<=n; j++)
        {
            if(dist[j]<min&&v[j]==0)
            {
                min = dist[j];
                point = j;
            }
        }
        v[point] =1;
        for(int j=1; j<=n; j++)
        {
            if(v[j]==0&&ma[point][j]!=0x3f3f3f3f&&dist[j]>ma[point][j]+dist[point])
                dist[j] = ma[point][j] + dist[point];
        }
    }
}
int main()
{
    int x, y, z;
    cin>>m>>n;
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            if(i==j)
                ma[i][j] = 0;
            else
                ma[i][j] = 0x3f3f3f3f;
        }
    }
    while(m--)
    {
        cin>>x>>y>>z;
        if(ma[x][y]>z)
            ma[x][y] = ma[y][x] = z;
    }
    Dijkstra(1);
    printf("%d\n", dist[n]);
    return 0;
}