文章作者：姜南(Slyar) 文章来源：Slyar Home (www.slyar.com) 转载请注明，谢谢合作。
Dijkstra(单源最短路径)算法的应用
Description
One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.
Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.
Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?
Input
Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
Output
Line 1: One integer: the maximum of time any one cow must walk.
Sample Input
4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3
Sample Output
10
Slyar:题目大意是给出一个有向图的带权邻接矩阵，然后给定目标节点X，求出其余结点出发到结点X再从X回到出发结点的最短路径和中最大的那个...=_=是不是很晕？晕的话去看英文...
恩，有向图，到X再从X回来，从X回来简单，一个Dijkstra就搞定，到X就比较麻烦了，一开始用Flyod做，结果超时...=_=后来发现只要把矩阵转置一下就好了，转置以后就又相当于从X出发了，再来一次Dijkstra搞定...很好理解的...
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#include <iostream> using namespace std; const int N = 1010; const int MAX = 1000001; int n, m; int map1[N][N]; int map2[N][N]; int dist1[N]; int dist2[N]; // dist[]记录节点x到其余各点的最短路径长度 void dijkstra(int map[][N], int dist[], int x) { int u, min; // 记录节点是否求解完毕 int visited[N] = {0}; // 节点x到自己的距离为0 dist[x] = 0; // 求解其余n-1个节点到x的最短路径 for (int i = 1; i <= n; i++) { // 选择与当前节点相邻且未访问的节点中距离最短的节点u for (int j = 1; j <= n; j++) { if (!visited[j] && dist[j] < min) { min = dist[j]; u = j; } } // 标记u已求解 visited[u] = 1; // 根据u来更新其余节点 for (int j = 1; j <= n; j++) { if (!visited[j] && dist[j] > dist[u] + map[u][j]) { dist[j] = dist[u] + map[u][j]; } } } } int main() { int ai, bi, ti, x; int ans = 0; cin >> n >> m >> x; //初始化 for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { map1[i][j] = MAX; map2[i][j] = MAX; } dist1[i] = MAX; dist2[i] = MAX; } //读入矩阵 for (int i = 1; i <= m; i++) { cin >> ai >> bi >> ti; map1[ai][bi] = ti; map2[bi][ai] = ti; } dijkstra(map1, dist1, x); dijkstra(map2, dist2, x); for (int i = 1; i <= n; i++) { if (ans < dist1[i] + dist2[i]) { ans = dist1[i] + dist2[i]; } } cout << ans << endl; return 0; } |
