Description
Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be able to use as little energy as possible to jump over the hurdles.
Obviously, it is not very difficult for a cow to jump over several very short hurdles, but one tall hurdle can be very stressful. Thus, the cows are only concerned about the height of the tallest hurdle they have to jump over.
The cows' practice room has N (1 ≤ N ≤ 300) stations, conveniently labeled 1..N. A set of M (1 ≤ M ≤ 25,000) one-way paths connects pairs of stations; the paths are also conveniently labeled 1..M. Path itravels from station Si to station Ei and contains exactly one hurdle of height Hi (1 ≤ Hi ≤ 1,000,000). Cows must jump hurdles in any path they traverse.
The cows have T (1 ≤ T ≤ 40,000) tasks to complete. Task i comprises two distinct numbers, Ai and Bi (1 ≤ Ai ≤ N; 1 ≤ Bi ≤ N), which
connote that a cow has to travel from station Ai to station Bi (by traversing over one or more paths over some route). The cows want to take a path the minimizes the height of the tallest hurdle they jump over when traveling
from Ai to Bi . Your job is to write a program that determines the path whose tallest hurdle is smallest and report that height.
Input
* Line 1: Three space-separated integers: N, M, and T
* Lines 2..M+1: Line i+1 contains three space-separated integers: Si , Ei , and Hi
* Lines M+2..M+T+1: Line i+M+1 contains two space-separated integers that describe task i: Ai and Bi
Output
* Lines 1..T: Line i contains the result for task i and tells the smallest possible maximum height necessary to travel between the stations. Output -1 if it is impossible to travel between the two stations.
Sample Input
5 6 3 1 2 12 3 2 8 1 3 5 2 5 3 3 4 4 2 4 8 3 4 1 2 5 1
Sample Output
4 8 -1
Source
USACO 2007 November Silver
題意：有一頭牛，要進行跳木樁訓練，已知有n個木樁，而且知道m對木樁之間的高度差。但是它很懶，它想盡可能的跳最小的高度就完成從任意一個木樁到任意一個木樁的跳躍，給m對點，問是否存在最小的跳躍高度使得其能夠完成跳躍，如果有就輸出最小高度；否則輸出-1。
解析：現在要記錄路徑“長度”，實際上是最大跳躍高度，說白了就是，這條路徑上所經過的相鄰兩木樁之間的差值的最大值，因為只要能跳過這個高度差最大的，高度差小的當然能跳過去了。由於是求任意兩木樁之間的最大高度差值的最小值，所以我們可以用Floyd算法，對其進行處理，處理之後得到的最終結果就是最大高度的最小值了。
AC代碼：
#include#include #include using namespace std; #define INF 123456789 int a[302][302]; //最大高度的最小值矩陣 int main(){ int n, m, t; int x, y, w; while(scanf("%d%d%d", &n, &m, &t)!=EOF){ for(int i=1; i<=n; i++) //初始化 for(int j=1; j<=n; j++) a[i][j] = i==j ? 0 : INF; for(int i=1; i<=m; i++){ //讀入高度差 scanf("%d%d%d", &x, &y, &w); a[x][y] = min(a[x][y], w); //更新最大高度差 } for(int k=1; k<=n; k++) //Floyd for(int i=1; i<=n; i++) for(int j=1; j<=n; j++){ a[i][j] = min(a[i][j], max(a[i][k], a[k][j])); } for(int i=1; i<=t; i++){ scanf("%d%d", &x, &y); printf("%d\n", a[x][y]==INF ? -1 : a[x][y]); //輸出，如果還是INF，那就代表不可達，兩者時之間沒有路徑滿足要求 } } return 0; }
