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];//記錄路徑數 
//*************************************************************
//Dijkstra-----數組實現
//單源最短路徑
//cost[i][j]為i,j間的距離
//lowcost[]————從點beg到其他點的最近距離
//path[]---------beg為根展開的樹，記錄父親結點
//*************************************************************
#define INF 0x3f3f3f3f  //這個無窮大不能太大，小心后面相加時溢出
#define typec int  //定義需要的數據類型
int path[MAXN],vis[MAXN];
void Dijkstra(typec cost[][MAXN],typec lowcost[MAXN],int n,int beg)
//結點是1~n標記的 
{
    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;//樹根的標記
    int pre;
    for(int num=2;num<n;num++)//決定從pre出發的n-1個最短路
    {
        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)//記憶性搜索，類似于動態規劃的方法，記錄下來 
{
    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));
               
    }    
}