Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To make things even nicer, his office is on one side of a forest, and his house is on the other. A nice walk through the forest, seeing the birds and chipmunks is quite enjoyable.
The forest is beautiful, and Jimmy wants to take a different route everyday. He also wants to get home before dark, so he always takes a path to make progress towards his house. He considers taking a path from A to B to be progress if there exists a route from B to his home that is shorter than any possible route from A. Calculate how many different routes through the forest Jimmy might take.
Input
Input contains several test cases followed by a line containing 0. Jimmy has numbered each intersection or joining of paths starting with 1. His office is numbered 1, and his house is numbered 2. The first line of each test case gives the number of intersections N, 1 < N ≤ 1000, and the number of paths M. The following M lines each contain a pair of intersections a b and an integer distance 1 ≤ d ≤ 1000000 indicating a path of length d between intersection a and a different intersection b. Jimmy may walk a path any direction he chooses. There is at most one path between any pair of intersections.
Output
For each test case, output a single integer indicating the number of different routes through the forest. You may assume that this number does not exceed 2147483647
Sample Input
5 6 1 3 2 1 4 2 3 4 3 1 5 12 4 2 34 5 2 24 7 8 1 3 1 1 4 1 3 7 1 7 4 1 7 5 1 6 7 1 5 2 1 6 2 1 0
Sample Output
2 4 题意： 给你一个图，找路。但是有个的条件：每走一步，如你选择要走a 到 b (如果a,b之间有路),那么必须保证a到终点的所有路 都小于 b到终点的每一条路。问满足这样的路径条数 有多少思路： 1. 1为起点，2为终点，因为要走ab路时，必须保证那个条件，所以从终点开始使用单源最短路Dijkstra算法，得到每个点到终点的最短路，保存在dis[]数组中。 2. 然后从起点开始深搜每条路，看看满足题意的路径有多少条。 3. 这样搜索之后，dp[1]就是从起点到终点所有满足题意的路径的条数。
1 #include <iostream>
2 #include <string.h>
3 using namespace std;
4 const int MAXN = 1234;
5 const int INF = 12345678;
6 int n, m;
7 int vis[MAXN];
8 int g[MAXN][MAXN];
9 int dis[MAXN];
10 int dp[MAXN];
11
12 void init(){
13 for(int i = 1; i <= n; i++){
14 for(int j = 1; j <= n; j++){
15 if(i == j)
16 g[i][j] = 0;
17 else g[i][j] = INF;
18 }
19 }
20 memset(vis,0,sizeof(vis));
21 memset(dp,-1,sizeof(dp));
22 }
23
24 void dij(int v0){
25 int pos = v0;
26 for(int i = 1; i <= n; i++){
27 dis[i] = g[v0][i];
28 }
29 vis[pos] = 1;
30 for(int i = 1; i < n; i++){
31 int mins = INF;
32 for(int j = 1; j <= n; j++){
33 if(!vis[j] && dis[j] < mins){
34 pos = j;
35 mins = dis[j];
36 }
37 }
38 vis[pos] = 1;
39 for(int j = 1; j <= n; j++){
40 if(!vis[j] && dis[j] > dis[pos] + g[pos][j])
41 dis[j] = dis[pos] + g[pos][j];
42 }
43 }
44 }
45
46 int dfs(int start){
47 int ans = 0;
48 if(dp[start] != -1)
49 return dp[start];
50 if(start == 2)
51 return 1;
52 for(int i = 1; i <= n; i++){
53 if(g[start][i] != INF && dis[start] > dis[i])
54 ans += dfs(i);
55 }
56 dp[start] = ans;
57 return dp[start];
58 }
59
60 int main(){
61 while(cin >> n && n){
62 init();
63 cin >> m;
64 for(int i = 1; i <= m; i++){
65 int a, b, w;
66 cin >> a >> b >> w;
67 g[a][b] = g[b][a] = w;
68 }
69 dij(2);
70 cout << dfs(1) << endl;
71 }
72 }