A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7733    Accepted Submission(s): 2851

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

【分析】

题目意思是说一个人要从上班的地方回到家里，途中会经过一些地方，按下面规则问他回家会有几种路线：

题目已经定义好1为上班地方2为家，每个地点之间的距离都已经知道，哪么如果从A到B的一条路，可以走的条件

是B到2（家）的路程必须小于从A到2（家）的路程，其实就是A到家的最短路径必须大于B到家的最短路径。

默认一定可以走到家，也就是一定会有一种路线。最后是让你计算有几种路线。可以用迪杰斯特拉算法找一下所有点到终点的最短路，然后DFS搜一下就行了。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
using namespace std;
typedef long long ll;
const int N=1005;
int n;//十字路口数
int w[1001][1001];
int dist[1001],dp[1001];
int s[1001];
void dijkstra(int v)//迪杰斯特拉算法
{
    int i,j,mins,index;
    for(i=1; i<=n; i++)
    {
        dist[i] = w[i][v];
        s[i] = 0;
    }
    dist[v] = 0;
    s[v] = 1;
    for(i=1; i<n; i++)
    {
        mins = 2000000;
        for(j=1; j<=n; j++)
        {
            if(s[j]==0 && dist[j]<mins)
            {
                mins = dist[j];
                index = j;
            }
        }
        if(mins == 2000000)//注意 若没有 会在中间溢出
            break;
        s[index] = 1;
        for(j=1; j<=n; j++)
        {
            if(s[j]==0 && dist[j]>dist[index]+w[j][index])
                dist[j] = dist[index]+w[j][index];
        }
    }
}

int dfs(int v)//记忆法深搜
{
    if(dp[v] != -1)
        return dp[v];
    if(v == 2)
        return 1;
    int i,temp,sum=0;
    for(i=1; i<=n; i++)
    {
        if(w[v][i]!=2000000 && dist[v] > dist[i])//有路相通而且要去的i点到终点站的距离要比v到终点站的距离小
        {
            temp = dfs(i);
            sum += temp;
        }
    }
    dp[v] = sum;
    return sum;
}

int main()
{
    while(cin>>n && n)
    {
        int i,j,d,m;
        cin>>m;
        for(i=1; i<=n; i++)
        {
            dp[i] = -1;
            for(j=1; j<=n; j++)
                w[i][j] = 2000000;
        }
        while(m--)
        {
            scanf("%d%d%d",&i,&j,&d);
            w[i][j] = w[j][i] = d;//无向图
        }
//求出各点到终点站的最短距离
        dijkstra(2);//2为终点站
        dfs(1);//从1出发
        cout<<dp[1]<<endl;
    }
    return 0;
}