Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1214    Accepted Submission(s): 394

本题的思想就是dfs+dijkstra。本题的意思是，如果在点B出存在一条路使得从B点出发可以比从A点出发跟快的到达home，所以B到home的最短距离要比A到home得最短距离小，所以本题首先要求出各点到home得距离，然后再用记忆搜索法搜索即可

代码：

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<queue>
  4  using namespace std;
  5  int dist[1005],dp[1005];
  6 typedef struct t
  7 {
  8     int end,len;
  9     struct t *next;
 10 }T;
 11 struct node
 12 {
 13     int data;
 14     int len;
 15     bool operator <(const node &a)const
 16     {
 17         return a.len<len;
 18     }
 19     T *next;
 20 }s[1005];
 21 void dijkstra(int v0)
 22 {
 23     s[v0].len=0;node cur;
 24     int visit[1005],mark=0;
 25     memset(visit,0,sizeof(visit));
 26     memset(dist,-1,sizeof(dist));
 27     priority_queue<node> qu;
 28     qu.push(s[v0]);
 29     dist[v0]=0;
 30     while(!qu.empty ())
 31     {
 32         cur=qu.top();
 33         while(visit[cur.data])
 34         {
 35             qu.pop();
 36             if(qu.empty ())
 37             {
 38                 mark=1;
 39                 break;
 40             }
 41             cur=qu.top();
 42         }
 43         if(mark)
 44             break;
 45         qu.pop();
 46         visit[cur.data]=1;
 47         T *p=cur.next;
 48         while(p)
 49         {
 50             if(!visit[p->end])
 51             {
 52                 if(dist[p->end]==-1||dist[cur.data]+p->len<dist[p->end] )
 53                     s[p->end].len=dist[p->end]=p->len+dist[cur.data];
 54                 qu.push (s[p->end]);
 55             }
 56             p=p->next;
 57         }
 58     }
 59 }
 60 int dfs(int v0)
 61 {
 62     int temp=0;
 63     if(v0==2)
 64         return 1;
 65     if(dp[v0]!=-1)
 66         return dp[v0];
 67     t *p=s[v0].next;
 68     while(p)
 69     {
 70         if(dist[v0]>dist[p->end])
 71         {
 72             dp[p->end]=dfs(p->end);
 73             temp+=dp[p->end];
 74         }
 75         p=p->next;
 76     }
 77     return temp;
 78 }
 79 int main()
 80 {
 81     int n,i,m,a,b,len;
 82     while(scanf("%d",&n)!=EOF)
 83     {
 84         if(n==0)
 85             break;
 86         scanf("%d",&m);
 87         T *p,*q;
 88         for(i=1;i<=n;i++)
 89         {
 90             s[i].next=NULL;
 91             s[i].data=i;
 92         }
 93         memset(dp,-1,sizeof(dp));
 94         for(i=1;i<=m;i++)
 95         {
 96             scanf("%d%d%d",&a,&b,&len);
 97             p=(T*)malloc(sizeof(T));
 98             p->end=b;
 99             p->len=len;
100             p->next=s[a].next;
101             s[a].next=p;
102             q=(T*)malloc(sizeof(T));
103             q->end=a;
104             q->len=len;
105             q->next=s[b].next;
106             s[b].next=q;
107         }
108         dijkstra(2);
109         printf("%d\n",dfs(1));
110     }
111     return 0;
112 }
113