HDU 1142 A Walk Through the Forest

这道题的题目描述对于英语不好的同学可能有点复杂，它的意思就是找到一条最短路，然后在保证最短路的前提下，找出有多长条路满足从1->2；也就是找出有多少条最短路。

首先，还是一个单源最短路来算出最短路的路程，然后搜索一下各条路就可以了，代码如下：

 

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4000    Accepted Submission(s): 1470

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 61 3 21 4 23 4 31 5 124 2 345 2 247 81 3 11 4 13 7 17 4 17 5 16 7 15 2 16 2 10

 

Sample Output

24
 
下面这份代码提交的时候要用C++，不要用G++，不然会RE
 
 
#include <iostream>#include <stdlib.h>#include <string.h>#include <algorithm>using namespace std;const int M = 1005;#define inf 15000000int n,m;int mp[M][M];bool visit[M];int a,b,c;int dis[M],dp[M];int Dijkstra( int st  ){    for( int i=0 ; i<=n ; i++ ){         visit[i] = false;         dis[i] = mp[st][i];    }    visit[st] = true;    dis[st] = 0;    int tmp,k;    for( int i=0 ; i<n ; i++ ){         tmp = inf;         for( int j=1 ; j<=n ; j++ ){              if( !visit[j] && tmp > dis[j] )                  tmp = dis[ k = j ];         }         visit[k] = true;         for( int j=1 ; j<=n ; j++ )              if( !visit[j] && dis[j] > dis[k]+mp[k][j] )                  dis[j] = dis[k] + mp[k][j];    }    return dis[1];}int DFS( int st ){    int sum = 0;    if( st == 2 )        return 1;    if( dp[st] != -1 )        return dp[st];    for( int i=1 ; i<=n ; i++ )         if( mp[st][i] != inf && dis[st] > dis[i] )             sum += DFS( i );    dp[st] = sum;    return dp[st];}int main(){    while( scanf("%d",&n) && n ){           scanf("%d",&m);           for( int i=1 ; i<=n ; i++ ){                dp[i] = -1;                for( int j=1 ; j<=n ; j++ ){                         mp[i][j] = inf;                }           }           for( int i=0 ; i<m ; i++ ){                scanf("%d%d%d",&a,&b,&c);                mp[a][b] = mp[b][a] = c;           }           int ans,res;           ans = Dijkstra( 2 );    //从终点算是为了计算各个点到终点的距离（dis【i】）            //printf("%d\n",ans);           res = DFS( 1 );           printf("%d\n",res);    }    return 0;}