# POJ2387 Til the Cows Come Home（最短路径）

## Til the Cows Come Home

Time Limit: 1000MSMemory Limit: 65536K
Total Submissions: 67136Accepted: 22570

### 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

90

### Hint

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

### Source

USACO 2004 November

### 链接

http://poj.org/problem?id=2387

### 代码

• Dijkstra
StatusAccepted
Time32ms
Memory4628kB
Length127
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 1010;
const int inf = 0x3f3f3f3f;
int G[maxn][maxn];
bool vis[maxn];
int dist[maxn];
int n, m;

void Dijkstra(int s)
{
for (int i = 1; i <= n; ++i)
{
dist[i] = inf;
vis[i] = 0;
}
dist[s] = 0;
for (int i = 1; i <= n; ++i)
{
int Min = inf, k = -1;
for (int j = 1; j <= n; ++j)
{
if (dist[j] < Min && !vis[j])
{
Min = dist[j];
k = j;
}
}
if (k == -1)
break;
vis[k] = true;
for (int j = 1; j <= n; ++j)
{
if (G[k][j] + Min < dist[j] && !vis[j])
dist[j] = Min + G[k][j];
}
}
}

int main()
{
while (scanf("%d%d", &m, &n) != EOF)
{
int u, v, w;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
G[i][j] = ((i == j) ? 0 : inf);
for (int i = 0; i < m; ++i)
{
scanf("%d%d%d", &u, &v, &w);
G[u][v] = G[v][u] = min(w, G[u][v]); // 处理重边
}
Dijkstra(1);
printf("%d\n", dist[n]);
}
return 0;
}
• SPFA
StatusAccepted
Time47ms
Memory764kB
Length1623
#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1010;
int n, m;
bool vis[maxn];
int dist[maxn];
int cnt[maxn];

struct Edge
{
int v, w;
Edge(int _v, int _w)
{
v = _v;
w = _w;
}
};

vector <Edge> edge[maxn];

void addedge(int u, int v, int w)
{
edge[u].push_back(Edge(v, w));
}

bool spfa(int s)
{
for (int i = 1; i <= n; ++i)
dist[i] = inf;
memset(vis, 0, sizeof(vis));
memset(cnt, 0, sizeof(cnt));
dist[s] = 0;
queue <int> q;
while (!q.empty())
q.pop();
q.push(s);
vis[s] = true;
cnt[s] = 1;
while (!q.empty())
{
int u = q.front();
vis[u] = false;
q.pop();
for (int i = 0; i < edge[u].size(); ++i)
{
int v = edge[u][i].v;
if (dist[v] > dist[u] + edge[u][i].w)
{
dist[v] = dist[u] + edge[u][i].w;
if (!vis[v])
{
vis[v] = true;
q.push(v);
if (++cnt[v] > n)
return false;
}
}
}
}
return true;
}

int main()
{
while (scanf("%d%d", &m, &n) != EOF)
{
int u, v, w;
for (int i = 0; i < m; ++i)
{
scanf("%d%d%d", &u, &v, &w);
}