2387. Til the Cows Come Home

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. The first line contains two integers t and n. Each line from 2 to t + 1 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. Print the minimum distance that Bessie must travel to get from landmark n to landmark 1.

Sample input	Sample output
5 5 1 2 20 2 3 30 3 4 20 4 5 20 1 5 100	90

РЕШЕНИЕ

графы - Дейкстра

Анализ алгоритма

При помощи алгоритма Дейкстры ищем кратчайший путь между вершинами 1 и n.

Реализация алгоритма

#include <cstdio>

#include <cstring>

#include <vector>

#define MAX 5010

#define INF 0x3F3F3F3F

using namespace std;

int i, t, n, b, e, w;

int used[MAX], dist[MAX];

vector<vector<pair<int, int> > > g;

void Relax(int v, int to, int cost)

{

if (dist[to] > dist[v] + cost)

dist[to] = dist[v] + cost;

}

int Find_Min(void)

{

int i, v, min = INF;

for(i = 1; i <= n; i++)

if (!used[i] && (dist[i] < min)) min = dist[i], v = i;

if (min == INF) return -1;

return v;

}

void Dijkstra(void)

{

memset(used,0,sizeof(used));

memset(dist,0x3F,sizeof(dist));

dist[1] = 0;

for(int i = 1; i < n; i++)

{

int v = Find_Min();

if (v == -1) break;

used[v] = 1;

for(int j = 0; j < g[v].size(); j++)

{

int to = g[v][j].first;

int cost = g[v][j].second;

if (!used[to]) Relax(v,to,cost);

}

}

}

int main(void)

{

scanf("%d %d",&t,&n);

g.resize(n+1);

for(i = 0; i < t; i++)

{

scanf("%d %d %d",&b,&e,&w);

g[b].push_back(make_pair(e,w));

g[e].push_back(make_pair(b,w));

}

Dijkstra();

printf("%d\n",dist[n]);

return 0;

}