HDU-3631 Shortest Path （floyd）

Description

When YY was a boy and LMY was a girl, they trained for NOI (National Olympiad in Informatics) in GD team. One day, GD team’s coach, Prof. GUO asked them to solve the following shortest-path problem.
There is a weighted directed multigraph G. And there are following two operations for the weighted directed multigraph:

(1) Mark a vertex in the graph.

(2) Find the shortest-path between two vertices only through marked vertices.

For it was the first time that LMY faced such a problem, she
was very nervous. At this moment, YY decided to help LMY to analyze the
shortest-path problem. With the help of YY, LMY solved the problem at
once, admiring YY very much. Since then, when LMY meets problems, she
always calls YY to analyze the problems for her. Of course, YY is very
glad to help LMY. Finally, it is known to us all, YY and LMY become
programming lovers.

Could you also solve the shortest-path problem?

 

Input

The input consists of multiple test cases. For each test case, the
first line contains three integers N, M and Q, where N is the number of
vertices in the given graph, N≤300; M is the number of arcs, M≤100000;
and Q is the number of operations, Q ≤100000. All vertices are number as
0, 1, 2, … , N - 1, respectively. Initially all vertices are unmarked.
Each of the next M lines describes an arc by three integers (x, y, c):
initial vertex (x), terminal vertex (y), and the weight of the arc (c).
(c > 0) Then each of the next Q lines describes an operation, where
operation “0 x” represents that vertex x is marked, and operation “1 x
y” finds the length of shortest-path between x and y only through marked
vertices. There is a blank line between two consecutive test cases.

End of input is indicated by a line containing N = M = Q = 0.

 

Output

Start each test case with "Case #:" on a single line, where # is the case number starting from 1.

For operation “0 x”, if vertex x has been marked, output “ERROR! At point x”.

For operation “1 x y”, if vertex x or vertex y isn’t marked,
output “ERROR! At path x to y”; if y isn’t reachable from x through
marked vertices, output “No such path”; otherwise output the length of
the shortest-path. The format is showed as sample output.

There is a blank line between two consecutive test cases.

Sample Input

5 10 10
1 2 6335
0 4 5725
3 3 6963
4 0 8146
1 2 9962
1 0 1943
2 1 2392
4 2 154
2 2 7422
1 3 9896
0 1
0 3
0 2
0 4
0 4
0 1
1 3 3
1 1 1
0 3
0 4
0 0 0
 
Sample Output

Case 1:
ERROR! At point 4
ERROR! At point 1
0
0
ERROR! At point 3
ERROR! At point 4
 
题目解析：每标记一个点就单独对这个点松弛。要注意有个 "There is a blank line between two consecutive test cases.”
 
代码如下：

 # include<iostream>
 # include<cstdio>
 # include<cstring>
 # include<queue>
 # include<algorithm>
 const int INF=<<;
 using namespace std;
 int mp[][];
 int n,m,q,mark[];
 void floyd(int k)
 {
     for(int i=;i<n;++i)
         for(int j=;j<n;++j)
             mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);
 }
 void work(int s,int t)
 {
     if(mark[s]==||mark[t]==)
         printf("ERROR! At path %d to %d\n",s,t);
     else{
         if(mp[s][t]!=INF)
             printf("%d\n",mp[s][t]);
         else
             printf("No such path\n");
     }
 }
 int main()
 {
     //freopen("Qcin.txt","r",stdin);
     int a,b,c,i,j,cas=;
     while(scanf("%d%d%d",&n,&m,&q)==)
     {
         if(n==&&m==&&q==)
             break;
         if(cas)
             printf("\n");
         for(i=;i<n;++i)
             for(j=;j<n;++j)
                 mp[i][j]=(i==j)?:INF;
         memset(mark,,sizeof(mark));
         while(m--)
         {
             scanf("%d%d%d",&a,&b,&c);
             mp[a][b]=min(mp[a][b],c);
         }
         int comd;
         printf("Case %d:\n",++cas);
         while(q--)
         {
             scanf("%d",&comd);
             if(comd==){
                 scanf("%d",&a);
                 if(mark[a]){
                     printf("ERROR! At point %d\n",a);
                 }else{
                     mark[a]=;
                     floyd(a);
                 }
             }
             else if(comd==){
                 scanf("%d%d",&a,&b);
                 work(a,b);
             }
         }
     }
     return ;
 }