HDU 1142 A Walk Through the Forest（最短路+dfs搜索）

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8850    Accepted Submission(s): 3267

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

一个人办公室在点1，家在点2，他要从办公室回家，他从点A到点B当且仅当从B到家的距离必任意一点从A到家的都小，求他回家路线的方案数

//最短路
#include <iostream> 
#include <algorithm> 
#include <cstring> 
#include <cstdio>
#include <vector> 
#include <queue> 
#include <cstdlib> 
#include <iomanip>
#include <cmath> 
#include <ctime> 
#include <map> 
#include <set> 
using namespace std; 
#define lowbit(x) (x&(-x)) 
#define max(x,y) (x>y?x:y) 
#define min(x,y) (x<y?x:y) 
#define MAX 100000000000000000 
#define MOD 1000000007
#define pi acos(-1.0) 
#define ei exp(1) 
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f 
#define mem(a) (memset(a,0,sizeof(a))) 
typedef long long ll;
const int N=1234;
ll g[N][N],vis[N],n,m,k,x,y,z;
ll dp[N],dis[N];
void init()
{
    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<i;j++)
        {
            g[i][j]=g[j][i]=INF;
        }
        g[i][i]=0;
    }
}
void dij(int x)
{
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)
    {
        dis[i]=g[x][i];
    }
    vis[x]=1;
    int v=x;
    int minn;
    for(int i=1;i<=n;i++)
    {
        minn=INF;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j] && minn>dis[j])
            {
                v=j;
                minn=dis[j];
            }
        }
        vis[v]=1;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]) dis[j]=min(dis[j],dis[v]+g[v][j]);
        }
    }
}
ll dfs(ll now)
{
    if(dp[now]>0) return dp[now];
    if(now==2) return 1;
    for(int i=1;i<=n;i++)
    {
        if(g[now][i]!=INF && dis[now]>dis[i])
            dp[now]+=dfs(i);
    }
    return dp[now];
}
int main()
{
    while(scanf("%lld",&n) && n)
    {
        init();
        scanf("%lld",&m);
        for(int i=0;i<m;i++)
        {
            scanf("%lld%lld%lld",&x,&y,&z);
            g[x][y]=g[y][x]=min(g[x][y],z);
        }
        dij(2);
        memset(dp,0,sizeof(dp));
        printf("%lld
",dfs(1));
    }
    return 0;
}