hdu1142 A Walk Through the Forest（dijkstra + 记忆化搜索）

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3018    Accepted Submission(s): 1101

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

 

题意：看样子很多人都把这题目看错了，以为是求最短路的条数。真正的意思是：假设 A 和 B 是相连的，当前在 A 处，如果 A 到终点的距离大于 B 到终点的距离，

则可以从 A 通往 B 处，问满足这种的条件的路径条数。

分析：1、以终点 2 为起点 dijkstra；

     2、直接DFS记忆化搜索。

 

#include<iostream>
#include<cstdio>
#include<string.h>
#define N 1010
#define INF 2000000000

using namespace std;

int map[N][N],lowcost[N],visited[N],d[N],p[N];


void dijkstra(int s,int n)
{
    memset(visited,false,sizeof(visited));
    int i,j,k,min;
    for(i=1;i<=n;i++)
    {
        lowcost[i]=map[s][i];
    }
    d[s]=0;
    visited[s]=true;
    for(i=1;i<n;i++)//我又写成 i<=n 了 调试很久！！！
    {
        min=INF;
        for(j=1;j<=n;j++)
        {
            if(!visited[j]&&min>lowcost[j])
            {
                min=lowcost[j];
                k=j;
            }
        }
        d[k]=min;
        visited[k]=true;
        for(j=1;j<=n;j++)
        {
            if(!visited[j]&&lowcost[j]>map[k][j]+d[k])
                lowcost[j]=map[k][j]+d[k];
        }
    }
}

int DFS(int s,int n)
{
    if(p[s]) return p[s];
    if(s==2)  return 1;
    int i,sum=0;
    for(i=1;i<=n;i++)
    {
        if(map[s][i]<INF&&d[s]>d[i])
        {
            if(p[i]) sum=sum+p[i];
            else sum=sum+DFS(i,n);
        }
    }
    sum=sum+p[s];
    p[s]=sum;
    return p[s];
}

int main()
{
    int i,j,n,m,u,v,w;
    while(cin>>n&&n)
    {
        cin>>m;
        memset(p,0,sizeof(p));
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                map[i][j]=(i==j?0:INF);
            }
        }
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            map[u][v]=map[v][u]=w;
        }
        dijkstra(2,n);
        cout<<DFS(1,n)<<endl;
    }
    return 0;
}