HDU3631:Shortest Path(Floyd)
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?
End of input is indicated by a line containing N = M = Q = 0.
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.
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
ERROR! At point 4
ERROR! At point 1
0
0
ERROR! At point 3
ERROR! At point 4
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; const int inf = 999999999; int n,m,k;
int map[305][305];
int hash[305]; void floyd(int k)
{
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 x,y,c,i,j,cas = 1;
while(~scanf("%d%d%d",&n,&m,&k),n+m+k)
{
memset(hash,0,sizeof(hash));
for(i = 0; i<=n; i++)
{
for(j = 0; j<=n; j++)
map[i][j] = inf;
map[i][i] = 0;
}
while(m--)
{
scanf("%d%d%d",&x,&y,&c);
if(c<map[x][y])
map[x][y] = c;
}
if(cas!=1)
printf("\n");
printf("Case %d:\n",cas++);
while(k--)
{
scanf("%d",&c);
if(c)
{
scanf("%d%d",&x,&y);
if(hash[x] && hash[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(hash[x])
printf("ERROR! At point %d\n",x);
else
{
hash[x] = 1;
floyd(x);//以新加入的点为中转点去更新最短路
}
}
}
} return 0;
}
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