POJ 2387 Til the Cows Come Home（Dijkstra算法）

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

题目大意：

给你一个数t,代表有几条已知的道路，再给出一个数n ,代表有n个点，让你求出从1到n的最短距离为多少

解题思路

标准迪杰斯特拉算法题目

算法思路过程如下：

参考代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define INF 0x3f3f3f3f
#define MME(a) memset(a,0,sizeof(a))
using namespace std;
int book[1005];//记录改点是否走过
int dp[1005];//记录1到任意节点的距离
int path[1005][1005];//记录两点之间的距离
int main()
{
int t, n;
while (~scanf("%d %d", &t, &n))
{
MME(book);
MME(dp);
MME(path);
for (int i = 1;i <= n;i++)
{
for (int j = 1;j <= n;j++)
{
if (i == j)//由于不能自环，所以主对角线置零
path[i][j] = 0;
else
path[i][j] = INF;
}
}
for (int i = 1;i <= t;i++)
{
int st, ed, len;
scanf("%d %d %d", &st, &ed, &len);
/*坑点：需要考虑两点重复输入更新*/
path[st][ed] = path[ed][st] = min(path[st][ed], len);
}
for (int i = 1;i <= n;i++)
{
dp[i] = path[1][i];/*dp数组存储1到任意节点的距离，如果没有关系就是正无穷*/
}
book[1] = 1;
int pos = 0;
for (int i = 1;i <= n-1;i++)
{
int min = INF;
for (int j = 1;j <= n;j++)
{
/*如果改点没走过，且1到该节点为最短距离时*/
if (!book[j] && dp[j] < min)
{
min = dp[j];
pos = j;
}
}
book[pos] = 1;
/*说明1到该节点的距离已为最短，然后标记上*/
for (int k = 1;k <= n;k++)
{
if (path[pos][k] != INF)/*如果pos到k有关系*/
{
/*判断1直接到k点和1点到pos点再由pos点到k点谁距离短，然后赋值*/
if (dp[k] > dp[pos] + path[pos][k])
dp[k] = dp[pos] + path[pos][k];
}
}
}
printf("%d\n", dp
);
}
}