hduoj----1142A Walk Through the Forest(记忆化搜索+最短路)

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5679    Accepted Submission(s): 2086

Problem Description

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

 

 

Source

University of Waterloo Local Contest 2005.09.24

 

这道题的大意为： 

                    这道题不同于以前简单的最短路径，它要求的是：如果A到B的过程中， 假如存在一个B点到家终点（2号点）的路径比所有A到家终点的路径要短。

                   即 distance[B][2]>distance[A][2];   那么A->B->2的路径算一条，计算这样的路径的条数.........

 对于这道题，我们其实可以这样思考

                                          我们不妨先算出。2号点到所有点的最短距离，然后再来优先搜索所有满足这样的条件的个数即为它的路径的个数.....

 

代码：

 1 #include<cstdio>

 2 #include<cstring>

 3 using namespace std;

 4 const int inf =0x3f3f3f3f;

 5 const int maxn =1005;

 6 bool vis[maxn];

 7 int lowc[maxn],map[maxn][maxn];

 8 int roadnum[maxn];

 9 void Dijkstra(int st,int n)

10 {

11     int minx;

12     memset(vis,0,sizeof(vis));

13     vis[st]=0;

14     for(int i=1;i<=n;i++){

15       lowc[i]=map[st][i];

16     }

17     lowc[st]=0;

18     int pre=st;

19     for(int i=1;i<n;i++){

20       minx=inf;

21       for(int j=1;j<=n;j++){

22         if(!vis[j]&&lowc[pre]+map[pre][j]<lowc[j]){

23             lowc[j]=lowc[pre]+map[pre][j];

24            }

25 

26         }

27         for(int j=1;j<=n;j++){

28           if(!vis[j]&&minx>lowc[j]){

29               minx=lowc[j];

30               pre=j;

31             }

32         }

33         vis[pre]=true;

34     }

35 }

36 

37 /*记忆化搜索dfs*/

38 int Dfs(int st,int n){

39 

40     if(st==2)  return 1;

41     else if(roadnum[st]) return roadnum[st] ;  //如果已经计算出来了就直接返回的路径条数

42      int sum=0;

43      for(int i=1;i<=n;i++){

44         if(map[st][i]!=inf&&lowc[st]>lowc[i])

45           sum+=Dfs(i,n);

46      }

47      roadnum[st]+=sum;

48      return roadnum[st];

49 }

50 void init(int n)

51 {

52   for(int i=1;i<=n;i++){

53     for(int j=i;j<=n;j++){

54         map[i][j]=map[j][i]=inf;

55         }

56     }

57 }

58 int main()

59 {

60     int n,m;

61     int x,y,val;

62     while(scanf("%d",&n)!=EOF&&n!=0){

63         scanf("%d",&m);

64         init(n);

65          memset(roadnum,0,sizeof(roadnum));

66         while(m--){

67           scanf("%d%d%d",&x,&y,&val);

68           if(map[y][x]>val)

69             map[y][x]=map[x][y]=val;

70         }

71         Dijkstra(2,n);

72         printf("%d\n",Dfs(1,n));

73     }

74   return 0;

75 }