hdu3631 floyd

Shortest Path

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3229    Accepted Submission(s): 767

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
   
   
   
   

 

Source

2010 Asia Regional Tianjin Site —— Online Contest

 

Recommend

lcy

 

看了别人的思路,说要用floyd 一开始不明白为什么可以标记一个更新一次floyd,看了floyd的思想（严蔚敏的数据结构那本）感觉此题就是为floyd而设的~~~具体看了就知道了

#include <iostream>
#include <cstdio>
#include <cstring>
#define inf 99999999
using namespace std;
int map[305][305];
int mark[305];
void floyd(int k,int n)
{
    int i,j;
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            if(map[i][j]>map[i][k]+map[k][j])
                map[i][j]=map[i][k]+map[k][j];
        }
    }
}
int main()
{
    int n,m,q;
    int cas=1;
    while(scanf("%d%d%d",&n,&m,&q))
    {
        if(n==0&&m==0&&q==0)
            break;
        int i,j;
        memset(mark,0,sizeof(mark));
        for(i=0;i<=n;i++)
            for(j=0;j<=n;j++)
                map[i][j]=(i==j?0:inf);
        for(i=0;i<m;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            if(w<map[u][v])
                map[u][v]=w;
        }
        if(cas!=1)
            printf("\n");
        printf("Case %d:\n",cas++);
        for(i=0;i<q;i++)
        {
            int f;
            scanf("%d",&f);
            if(f==0)
            {
                int t;
                scanf("%d",&t);
                if(mark[t])
                    printf("ERROR! At point %d\n",t);
                else
                {
                    mark[t]=1;
                    floyd(t,n);
                }
            }
            else
            {
                int u,v;
                scanf("%d%d",&u,&v);
                if(!mark[u]||!mark[v])
                    printf("ERROR! At path %d to %d\n",u,v);
                else if(map[u][v]>=inf)
                    printf("No such path\n");
                else
                    printf("%d\n",map[u][v]);
            }
        }
    }
    return 0;
}