另一個 stay walk mini sts 題意 scanf rom ted

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. 題意 ： 有 N 個點， T組邊的關系，求第一個點到第N 個點的最小距離。 註意 ： 有個坑點啊，就是從一個點到另一個點可能有多種路徑到達 ！！ Dijkstra算法 ：

const int inf = 1e5+5;  // 這個值要註意，避免越屆
int edge[1005][1005];
int t, n, ans;
int d[1005];
bool used[1005];

void dij(){
    for(int i = 1; i <= n; i++){
        d[i] = inf;
    }    
    memset(used, false, sizeof(used));
    d[1] = 0;
    while(1){
        int v = -1;
        for(int i = 1; i <= n; i++){
            if (!used[i] && (v == -1 || d[i] < d[v])) v = i;
        }
        if (v == -1) break;
        used[v] = true;
        for(int i = 1; i <= n; i++){
            d[i] = min(d[i], d[v]+edge[v][i]);
        }
    }    
}

int main() {
    int a, b, c;
    
    while(~scanf("%d%d", &t, &n)){
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                if (i == j) edge[i][j] = 0;
                else edge[i][j] = inf;
            }
        }
        for(int i = 1; i <= t; i++){
            scanf("%d%d%d", &a, &b, &c);
            if (edge[a][b] > c)  // 尋求最短的邊保存起來
                edge[a][b] = edge[b][a] = c;
        }
        dij();
        printf("%d\n", d[n]);
    }

    return 0;
}

最短路 - dijkstra