最短路和次短路问题,dijkstra算法

  1 /* 
  2  *题目大意: 
  3  *在一个有向图中,求从s到t两个点之间的最短路和比最短路长1的次短路的条数之和; 
  4  * 
  5  *算法思想: 
  6  *用A*求第K短路,目测会超时,直接在dijkstra算法上求次短路; 
  7  *将dist数组开成二维的,即dist[v][2],第二维分别用于记录最短路和次短路; 
  8  *再用一个cnt二维数组分别记录最短路和次短路的条数; 
  9  *每次更新路径的条数时,不能直接加1,,应该加上cnt[u][k],k为次短路径或者最短路径的标记; 
 10  *图有重边,不能用邻接矩阵存储; 
 11  *不知道为什么,题目上说的是N and M, separated by a single space, with 2≤N≤1000 and 1 ≤ M ≤ 10000; 
 12  *而我的代码硬是把N开成1W了才过,求解释,RE了无数次,擦; 
 13 **/ 
 14 #include<iostream>  
 15 #include<cstdio>  
 16 #include<cstring>  
 17 #include<string>  
 18 #include<algorithm>  
 19 using namespace std;  
 20    
 21 const int N=11111 22 const int M=111111 23 const int INF=0xffffff 24    
 25 struct node  
 26  27     int to;  
 28     int w;  
 29     int next;  
 30 };  
 31    
 32 node edge[N];  
 33 int head[N];  
 34    
 35 int dist[N][2],cnt[N][2];  
 36 bool vis[N][2];  
 37 int n,m,s,t,edges;  
 38    
 39 void addedge(int u,int v,int w)  
 40  41     edge[edges].w=w;  
 42     edge[edges].to=v;  
 43     edge[edges].next=head[u];  
 44     head[u]=edges++ 45  46    
 47 int dijkstra()  
 48  49     int k;  
 50     for(int i=0; i<=n; i++ 51     {  
 52         dist[i][0]=dist[i][1]=INF;  
 53         vis[i][0]=vis[i][1]=0 54         cnt[i][0]=cnt[i][1]=0 55     }  
 56     cnt[s][0]=1,dist[s][0]=0 57    
 58     for(int i=1; i<=n*2; i++ 59     {  
 60         int u=-1 61         int min_dist=INF;  
 62         for(int j=1; j<=n; j++ 63             for(int flag=0; flag<2; flag++ 64                 if(!vis[j][flag]&&dist[j][flag]<min_dist)  
 65                 {  
 66                     min_dist=dist[j][flag];  
 67                     u=j;  
 68                     k=flag;  
 69                 }  
 70         if(u==-1 71             break 72         vis[u][k]=true 73         for(int e=head[u]; e!=-1; e=edge[e].next)  
 74         {  
 75             int j=edge[e].to;  
 76             int tmp=dist[u][k]+edge[e].w;  
 77    
 78             if(tmp<dist[j][0])//tmp小于最短路径长:  
 79             {  
 80                 dist[j][1]=dist[j][0];//次短路径长  
 81                 cnt[j][1]=cnt[j][0];//次短路径计数  
 82                 dist[j][0]=tmp;//最短路径长  
 83                 cnt[j][0]=cnt[u][k];//最短路径计数  
 84             }  
 85    
 86             else if(tmp==dist[j][0])//tmp等于最短路径长:  
 87             {  
 88                 cnt[j][0]+=cnt[u][k];//最短路径计数  
 89             }  
 90    
 91             else if(tmp<dist[j][1])//tmp大于最短路径长且小于次短路径长:  
 92             {  
 93                 dist[j][1]=tmp;//次短路径长  
 94                 cnt[j][1]=cnt[u][k];//次短路径计数  
 95             }  
 96    
 97             else if(tmp==dist[j][1])//tmp等于次短路径长:  
 98             {  
 99                 cnt[j][1]+=cnt[u][k];//次短路径计数  
100             }  
101         }  
102     }  
103    
104     int res=cnt[t][0];  
105     if(dist[t][0]+1==dist[t][1])//判断最短路和次短路是否相差1  
106         res+=cnt[t][1];  
107     return res;  
108 109    
110 int main()  
111 112     //freopen("C:\\Users\\Administrator\\Desktop\\kd.txt","r",stdin);  
113     int tcase;  
114     scanf("%d",&tcase);  
115     while(tcase--116     {  
117         edges=0118         scanf("%d%d",&n,&m);  
119         memset(head,-1,sizeof(head));  
120         int u,v,w;  
121         for(int i=0; i<m; i++122         {  
123             scanf("%d%d%d",&u,&v,&w);  
124             addedge(u,v,w);  
125         }  
126         scanf("%d%d",&s,&t);  
127         printf("%d\n",dijkstra());  
128     }  
129     return 0130

 

posted @ 2017-08-10 07:42  Hammer_cwz_77  阅读(2090)  评论(0编辑  收藏  举报