hdu 3631 Shortest Path（floyd）

Shortest Path

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5061    Accepted Submission(s): 1225

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

 

题意：一个单向图，先输入n个点，m条边，k次询问。接下来m行输入相连点的边和距离，然后输入k次询问。输入0，x，表示标记x点，输入1，x，y，表示询问点x到点y的距离。只能走已标记的点。如果标记的点已标记 输出 ERROR! At point x，如果询问距离的两点有未标记的，输出 ERROR! At path x to y，如果都标记了，但是走不通，则输出 No such path，否则输出两点之间的距离。

 

floyd的灵活运用，直接用新标记的点更新，避免了O(n*n*n)

 

附上代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>
#define INF 0x3f3f3f3f
using namespace std;
int n,m,k;
int map[305][305];
int vis[305];
void floyd(int k)  ///新插入的点
{
    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 cas=1,i,j;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        if(n==0&&m==0&&k==0)
        break;
        memset(vis,0,sizeof(vis));
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
                if(i==j) map[i][j]=0;
                else     map[i][j]=INF;
        int a,b,c;
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(map[a][b]>c)
                map[a][b]=c;
        }
        if(cas!=1)
            printf("\n");
        printf("Case %d:\n",cas++);
        int x,y;
        while(k--)
        {
            scanf("%d",&c);
            if(c)
            {
                scanf("%d%d",&x,&y);
                if(vis[x]&&vis[y])
                {
                    if(map[x][y]!=INF)
                        printf("%d\n",map[x][y]);
                    else
                        printf("No such path\n");
                }
                else
                    printf("ERROR! At path %d to %d\n",x,y);
            }
            else
            {
                scanf("%d",&x);
                if(vis[x])
                    printf("ERROR! At point %d\n",x);
                else
                {
                    vis[x]=1;
                    floyd(x);
                }
            }
        }
    }
    return 0;
}