USACO 2008 March Gold cowjog

Calculate the Kth shortest path in a Directed Acyclic Graph(DAG). Although there exists a complicated way to calculate the Kth shortest Path in Arbitrary Graph, we can get the answer using DP by taking advantage of the property of DAG. A DP way is available because the DAG naturally has got an order which we use frequently.

 

We just maintain an array of size K for each node,  representing first K shortest path from source to that node. When we computing a node I, we just consider all the nodes that leads an edge to I. All we need to do is merging the arrays to get the first K items, then we fill them in the array of Node I.

 

Note: there may be multiple edges between two nodes.

 

Here is the code:

1. #include <iostream>
3. using namespace std;
5. const int N = 1001;
7. int ad[N][N], len[N][N], d[N][110];
8. int pt[N];
9. int n, m, kth;
11. int main()
12. {
13.     int i, j, k, t;
14.     
15.     freopen("cowjog.in", "r", stdin);
16.     freopen("cowjog.out", "w", stdout);
17.     
18.     scanf("%d%d%d",&n, &m, &kth);
19.     for(; m; m--)
20.     {
21.         scanf("%d%d%d",&i, &j, &k);
22.         ad[j][++ad[j][0]] = i;
23.         len[j][ad[j][0]] = k;
24.     }
26.     memset(d, -1, sizeof(d));
27.     
28.     d[n][0] = 0;
29.     for(i = n - 1; i >= 1; i--)
30.     {
31.         memset(pt, 0, sizeof(pt));
32.         
33.         for(int cnt = 0; cnt < kth; cnt++)
34.         {
35.             int tmp = -1, minD = INT_MAX;
36.             for(j = 1; j <= ad[i][0]; j++)
37.             {
38.                 k = ad[i][j], t = len[i][j];
39.                 if(d[k][pt[j]] != -1 && t + d[k][pt[j]] < minD)
40.                 {
41.                     minD = t + d[k][pt[j]];
42.                     tmp = j;
43.                 }
44.             }
45.             if(minD != INT_MAX)
46.             {
47.                 d[i][cnt] = minD;
48.                 pt[tmp]++;
49.             }
50.         }
51.     }
52.     for(i = 0; i < kth; i++)
53.         printf("%d/n",d[1][i]);
56.     scanf("%d",&n);
57.     return 0;
58. }