hdu 1142 A Walk Through the Forest（dijkstra）

http://acm.hdu.edu.cn/showproblem.php?pid=1142

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

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，就得到了最短的一条路，作为找路的最低限度。

2.然后深搜每条路，看看满足题意的路径有多少条。当然，这个需要从起点开始搜，因为dis[i]数组中保存的都是该点到终点的最短距离。

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。

#include<iostream>
using namespace std;
#define INF 9999999
#define MAX 1001
int map[MAX][MAX];
int dis[MAX],v[MAX],sum[MAX];
void dijkstra(int s,int n)
{
int i,j,min,k;
for(i=1;i<=n;i++)
  dis[i]=map[s][i];
memset(v,0,sizeof(v));
v[s]=1;
dis[s]=0;
for(i=2;i<=n;i++)
{
  k=s;
  min=INF;
  for(j=1;j<=n;j++)
   if(!v[j]&&min>dis[j])
   {
    min=dis[j];
    k=j;
   }
  v[k]=1;
  for(j=1;j<=n;j++)
   if(!v[j]&&dis[j]>dis[k]+map[k][j])
    dis[j]=dis[k]+map[k][j];
}
}
int dfs(int s,int n)
{
int i;
if(sum[s])
  return sum[s];
if(s==2)
  return 1;
for(i=1;i<=n;i++)
  if(map[s][i]!=INF)
  {
   if(dis[s]>dis[i])
    sum[s]+=dfs(i,n);
  }
return sum[s];
}
int main()
{
int n,m;
while(~scanf("%d",&n),n)
{
  scanf("%d",&m);
  int a,b,q;
  int i,j;
  for(i=1;i<=n;i++)
   for(j=1;j<=n;j++)
    map[i][j]=INF;
  while(m--)
  {
   scanf("%d%d%d",&a,&b,&q);
   if(map[a][b]==INF||map[a][b]>q)
    map[a][b]=map[b][a]=q;
  }
  dijkstra(2,n);
  memset(sum,0,sizeof(sum));
  printf("%d\n",dfs(1,n));
}
return 0;
}