体会：这个题做了好久，还是参考了别人的代码。不过体会到不能随便用dfs了。

关键：用DIJKTRA,不过不知道具体是怎么样，，不过感觉这个很靠谱

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

using namespace std;


#define N 1001

int map[N][N];
int dis[N];
bool vist[N];
int n;
int m;
int ans;
int g[N];

void dfs(int k)
{
    cout<dis[i]+map[i][k])
            dis[k]=dis[i]+map[i][k];
    }
}

void put()
{
    int i,j,k;
    for (i=1;i<=n;i++)
    {
        for (j=1;j<=n;j++)
            cout<map[i][j]+dis[j])
                dis[i]=map[i][j]+dis[j];
        }
    }
}

int get(int k)  //不能随便用dfs
{
    int i,j;
    if (g[k])
        return g[k];
    if (k==2)
        return 1;
    for (i=1;i<=n;i++)
    {
        if (map[k][i]&&dis[k]>dis[i])
            g[k]+=get(i);
    }
    return g[k];
}

int main()
{
    int i,j,k;
    while (cin>>n)
    {
        if (n==0)
            break;
        cin>>m;
        for (i=1;i<=n;i++)
            for (j=1;j<=n;j++)
                map[i][j]=0;
        while (m--)
        {
            cin>>i>>j>>k;
            map[i][j]=map[j][i]=k;
        }
        for (i=1;i<=n;i++)
            dis[i]=100000000;
        dis[2]=0;
        memset(vist,false,sizeof(vist));
        //dfs(1);  //介个不行啊，，，这只能找到一点到另一点之间的最短距离，，不好用。

         dij();

        ans=0;
        //make(1);
        //put();
        memset(g,0,sizeof(g));
        ans=get(1);
        cout< 
 
 
 
 
 

 
 
 
 
A Walk Through the Forest 
 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
 Total Submission(s): 2856    Accepted Submission(s): 1034
 
 
 
 
 
 

   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  
 
 
 

     
 
 
 
 
 

   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 
 
 
 

     
 
 
 
 
 

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