A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2121    Accepted Submission(s): 756


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
 

Source
University of Waterloo Local Contest 2005.09.24
 

Recommend
Eddy
 
 
 
  
#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define MAXN 1005
int cost[MAXN][MAXN];
int dis[MAXN];
int sum[MAXN];//记录路径数 
//*************************************************************
//Dijkstra-----数组实现
//单源最短路径
//cost[i][j]为i,j间的距离
//lowcost[]————从点beg到其他点的最近距离
//path[]---------beg为根展开的树，记录父亲结点
//*************************************************************
#define INF 0x3f3f3f3f  //这个无穷大不能太大，小心后面相加时溢出
#define typec int  //定义需要的数据类型
int path[MAXN],vis[MAXN];
void Dijkstra(typec cost[][MAXN],typec lowcost[MAXN],int n,int beg)
//结点是1~n标记的 
{
    int i,j;
    typec minc;
    memset(vis,0,sizeof(vis));
    vis[beg]=1;
    for(i=1;i<=n;i++)
    {
        lowcost[i]=cost[beg][i];path[i]=beg;
    }    
    lowcost[beg]=0;
    path[beg]=-1;//树根的标记
    int pre;
    for(int num=2;num<n;num++)//决定从pre出发的n-1个最短路
    {
        minc=INF;
        for(j=1;j<=n;j++)
           if(vis[j]==0&&lowcost[j]<minc)
           {pre=j;minc=lowcost[j];}
        if(minc>=INF)break;
        vis[pre]=1;
        for(j=1;j<=n;j++)
          if(vis[j]==0&&lowcost[pre]+cost[pre][j]<lowcost[j])
             {lowcost[j]=lowcost[pre]+cost[pre][j];path[j]=pre;}
    }     
}
//**********************************************************************   
int dfs(int i,int n)//记忆性搜索，类似于动态规划的方法，记录下来 
{
    if(i==2)  return 1;
    if(sum[i]!=-1)  return sum[i];
    int cnt=0;
    for(int j=1;j<=n;j++)
    {
        if(cost[i][j]<INF&&dis[j]<dis[i])
           cnt+=dfs(j,n);
    }
    sum[i]=cnt; 
    return sum[i];   
}    
int main()
{
    int i,j;
    int n,m;
    int a,b,d;
    while(scanf("%d",&n),n)
    {
        scanf("%d",&m);
        for(i=1;i<=n;i++)
           for(j=1;j<=n;j++)
           {
               if(i==j)cost[i][j]=0;
               else cost[i][j]=INF;
           } 
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&d);
            cost[a][b]=d;
            cost[b][a]=d;
        }
        Dijkstra(cost,dis,n,2);
        memset(sum,-1,sizeof(sum));
        printf("%d\n",dfs(1,n));
               
    }    
}