POJ 2387 Til the Cows Come Home

题目描述：

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.

　　　　这是道模版题，给出边数Ｔ和某一点Ｎ，求起点到Ｎ点的最短路径。我们可以把Ｎ点看成终点。应用Ｄｉｊｋｓｔｒａ算法求最短路即可。
　　　　不过，要注意处理重边问题，我刚开始就是没想到重边，wrong answer到怀疑人生。。。。
　　　　

代码：

#include <cstdio>#include <cstring>const int MAX = 2000 + 10;const int INF = 0xFFFFFFE;int a[MAX][MAX], dist[MAX];int T, N;bool book[MAX];void Dijkstra(){    memset(book, false, sizeof book);    for (int i = 1; i <= N; i++) dist[i] = a[1][i];//初始化dist数组    book[1] = true;    for (int i = 1; i < N; i++) {        int Min = INF, u = 0;        for (int j = 1; j <= N; j++) {//找出当前节点能到达的最短路径的节点u            if (!book[j] && Min > dist[j]) {                Min = dist[j];                u = j;            }        }        book[u] = true;//当前节点置为u        for (int k = 1; k <= N; k++) {//松弛边            if (!book[k] && dist[k] > dist[u] + a[u][k])dist[k] = dist[u] + a[u][k];        }    }}int main(){    while (scanf("%d%d", &T, &N) == 2) {        for (int i = 1; i <= N; i++) {//初始化图            a[i][i] = 0;            for (int j = 1; j <= N; j++)                if (i != j) a[i][j] = a[j][i] =  INF;        }        int x = 0, y = 0, w = 0;        for (int j = 1; j <= T; j++) {            scanf("%d%d%d", &x, &y, &w);            if (a[x][y] > w)a[x][y] = a[y][x] = w;//处理重边问题        }        Dijkstra();        printf("%d\n", dist[N]);    }    return 0;}