hdoj 3631 Shortest Path

 

Shortest Path

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1386    Accepted Submission(s): 336

Problem 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
   
   
   
   
   
   
   
   

    
    
    
    这是我在hdoj做的第一道题，以后要常来，不能只去poj。
   
   
   
   
   
   
   
   

    
    
    
    说一下算法：这道题是变相的Floyd算法，开始我写了一个Floyd函数，o（n^3），
   
   
   
   
   
   
   
   

    
    
    
    当然会TLE。这里的Floyd，中间的那个点由输入提供，就不要写三重循环了枚举他了
   
   
   
   

 

#include<iostream> using namespace std; #define inf 0xfffffff//7个f，若是8个f ，后面a[ii][jj]>a[ii][t]+a[t][jj] // 判断中a[ii][t]+a[t][jj]的和会溢出变为负值 int main() { int a[400][400],mark[400];//它说N<=300，可开300就会RE，不知道为什么 int n,m,q; int cnt=0; while(scanf("%d%d%d",&n,&m,&q)==3&&!(n==0&&m==0&&q==0)) {//m,q可以为0的，在这wa了好多次 cnt++; if(cnt!=1)printf("/n"); printf("Case %d:/n",cnt);//这个换行一定要判断，且不能放到末尾，否则会多输出空行 for(int i=0;i<n;i++) { for(int j=0;j<n;j++) { if(i==j)a[i][j]=0; else a[i][j]=inf; } mark[i]=0; } for(int i=0;i<m;i++) { int v,u,w; scanf("%d%d%d",&v,&u,&w); a[v][u]=min(a[v][u],w); } for(int i=0;i<q;i++) { int op; scanf("%d",&op); if(op==0) { int t; scanf("%d",&t); if(mark[t]) { printf("ERROR! At point %d/n",t); } else { mark[t]=1; for(int ii=0;ii<n;ii++) { for(int jj=0;jj<n;jj++) { if(a[ii][jj]>a[ii][t]+a[t][jj]) { a[ii][jj]=a[ii][t]+a[t][jj]; } } } } } else { int v,u; scanf("%d%d",&v,&u); if(!mark[v]||!mark[u]) { printf("ERROR! At path %d to %d/n",v,u); } else if(a[v][u]>=inf) { printf("No such path/n"); } else { printf("%d/n",a[v][u]); } } } } }

2010-09-1610:07:32