Problem J

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. <br>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. <br>

 

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. <br>

 

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<br>

 

Sample Input

5 61 3 21 4 23 4 31 5 124 2 345 2 247 81 3 11 4 13 7 17 4 17 5 16 7 15 2 16 2 10

 

Sample Output

24

 简单题意：

  吉米在回家的路上，路过森林是一件非常惬意的事情。现在需要编写一个程序，求出吉米穿越森林回家会有多少不同的方法。

解题思路形成过程：

  这是一个典型的求出最短路径的问题。用到了课堂上最后讲到的一种实用算法，Spfa和深度搜索。套入模板，代码就能一步一步完成。

感想：

  图论方面，算法如果自己想，难度是非常大的。用上课堂上的算法，所有题目简单了许多。

AC代码：

#include<iostream>
#include<queue>
using namespace std;
 
typedef struct n1
{
    int  distens,flog;
}node;
node N[1005];
int map[1005][1005],k;
int direct[1005];
void set(int n)
{
    int i,j,m,n1,n2,d;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
        {
            map[i][j]=-1;
        }
         N[i].distens=10000000;N[i].flog=0;direct[i]=0;
    }
 
    scanf("%d",&m);
    while(m--)
    {
        scanf("%d%d%d",&n1,&n2,&d);
        if(map[n1][n2]!=0||map[n1][n2]>d)
        map[n1][n2]=map[n2][n1]=d;
    }
}
void spfa(int n)
{
    queue<int> Q;
    int now;
    int i;
     N[2].distens=0;N[2].flog=1;
    Q.push(2);
    while(!Q.empty())
    {
        now=Q.front();
        Q.pop();
        N[now].flog=0;
        for(i=1;i<=n;i++)
        if(map[now][i]!=-1)
        {
            if(N[i].distens>N[now].distens+map[now][i])
            {
                N[i].distens=N[now].distens+map[now][i];
                if(N[i].flog==0)
                {
                    N[i].flog=1;
                    Q.push(i);
                }
            }
        }
    }
}
int DFS(int now,int n)
{
    int i;
    if(direct[now]>0)
    return direct[now];
    if(now==2)
    {
        return 1;
    }
    for(i=1;i<=n;i++)
    if(map[now][i]!=-1&&N[now].distens>N[i].distens)
    {
        direct[now]+=DFS(i,n);
    }
 
   return direct[now];
}
int main()
{
    int n;
    while(scanf("%d",&n)>0&&n)
    {
        set(n);
        spfa(n);
        k=DFS(1,n);
        printf("%d\n",k);
    }
}