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This is a explanation of this problem from USACO's training website. I have converted it to markdown. Please do not just copy code; you will not learn anything; at least type it out and understand so you can do it yourself in the future!
It’s dinner time, and the cows are out in their separate pastures. Farmer John rings the bell so they will start walking to the barn. Your job is to figure out which one cow gets to the barn first (the supplied test data will always have exactly one fastest cow).
Between milkings, each cow is located in her own pasture, though some pastures have no cows in them. Each pasture is connected by a path to one or more other pastures (potentially including itself). Sometimes, two (potentially self-same) pastures are connected by more than one path. One or more of the pastures has a path to the barn. Thus, all cows have a path to the barn and they always know the shortest path. Of course, cows can go either direction on a path and they all walk at the same speed.
The pastures are labeled ‘a’..’z’ and ‘A’..’Y’. One cow is in each pasture labeled with a capital letter. No cow is in a pasture labeled with a lower case letter. The barn’s label is ‘Z’; no cows are in the barn, though.
PROGRAM NAME: comehome
INPUT FORMAT
| Lines 1: | Integer P (1 <= P <= 10000) the number of paths that interconnect the pastures (and the barn) |
| Lines 2..P+1: | Space separated, two letters and an integer: the names of the interconnected pastures/barn and the distance between them (1 <= distance <= 1000) |
SAMPLE INPUT (file comehome.in)
5
A d 6
B d 3
C e 9
d Z 8
e Z 3
OUTPUT FORMAT
A single line containing two items: the capital letter name of the pasture of the cow that arrives first back at the barn, the length of the path followed by that cow.
SAMPLE OUTPUT (file comehome.out)
B 11
CODE
Java
C++
Pascal
ANALYSIS
Russ Cox
We use the Floyd-Warshall all pairs shortest path algorithm to calculate the minimum distance between the barn and all other points in the pasture. Then we scan through all the cow-containing pastures looking for the minimum distance.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#define INF 60000 /* bigger than longest possible path */
int dist[52][52];
int
char2num(char c)
{
assert(isalpha(c));
if(isupper(c))
return c-'A';
else
return c-'a'+26;
}
void
main(void)
{
FILE *fin, *fout;
int i, j, k, npath, d;
char a, b;
int m;
fin = fopen("comehome.in", "r");
fout = fopen("comehome.out", "w");
assert(fin != NULL && fout != NULL);
for(i=0; i<52; i++)
for(j=0; j<52; j++)
dist[i][j] = INF;
for(i=0; i<26; i++)
dist[i][i] = 0;
fscanf(fin, "%d\n", &npath);
for(i=0; i<npath; i++) {
fscanf(fin, "%c %c %d\n", &a, &b, &d);
a = char2num(a);
b = char2num(b);
if(dist[a][b] > d)
dist[a][b] = dist[b][a] = d;
}
/* floyd warshall all pair shortest path */
for(k=0; k<52; k++)
for(i=0; i<52; i++)
for(j=0; j<52; j++)
if(dist[i][k]+dist[k][j] < dist[i][j])
dist[i][j] = dist[i][k]+dist[k][j];
/* find closest cow */
m = INF;
a = '#';
for(i='A'; i<='Y'; i++) {
d = dist[char2num(i)][char2num('Z')];
if(d < m) {
m = d;
a = i;
}
}
fprintf(fout, "%c %d\n", a, m);
exit(0);
}
Wouter Waalewijn
This analysis of and code is by Wouter Waalewijn of The Netherlands.
When looking at the problem the first thing you can conclude is that for the solution you will need to know all the distances from the pastures to the barn. After calculating them you only have to check all these distances and pick out the nearest pasture with a cow in it, and that’s all.
Because the amount of vertices (=pastures+barn) is small, running Floyd/Warshall algorithm will solve the problem easily in time. If you think programming Floyd/Warshall is easier than Dijkstra, just do it. But you can also solve the problem running Dijkstra once, which of course speeds up your program quite a bit. Just initialise the barn as starting point, and the algorithm will find the distances from the barn to all the pastures which is the same as the distances from all the pastures to the barn because the graph is undirected. Using dijkstra for the solution would make far more complex data solvable within time. Here below you can see my implementation of this solution in Pascal. It might look big, but this way of partitioning your program keeps it easy to debug.
Var Dist:Array [1..58] of LongInt; {Array with distances to barn}
Vis :Array [1..58] of Boolean; {Array keeping track which
pastures visited}
Conn:Array [1..58,1..58] of Word; {Matrix with length of edges, 0 = no edge}
Procedure Load;
Var TF :Text;
X,D,E:Word;
P1,P2:Char;
Begin
Assign(TF,'comehome.in');
Reset(TF);
Readln(TF,E); {Read number of edges}
For X:=1 to E do
Begin
Read(TF,P1); {Read both pastures and edge
length}
Read(TF,P2);
Read(TF,P2); {Add edge in matrix if no edge between P1 and P2 yet or}
Readln(TF,D); {this edge is shorter than the shortest till now}
If (Conn[Ord(P1)-Ord('A')+1,Ord(P2)-Ord('A')+1]=0) or
(Conn[Ord(P1)-Ord('A')+1,Ord(P2)-Ord('A')+1]>D) then
Begin
Conn[Ord(P1)-Ord('A')+1,Ord(P2)-Ord('A')+1]:=D;
Conn[Ord(P2)-Ord('A')+1,Ord(P1)-Ord('A')+1]:=D;
End;
End;
Close(TF);
For X:=1 to 58 do
Dist[X]:=2147483647; {Set all distances to infinity}
Dist[Ord('Z')-Ord('A')+1]:=0; {Set distance from barn to barn to 0}
End;
Procedure Solve;
Var X,P,D:LongInt; {P = pasture and D = distance}
Begin
Repeat
P:=0;
D:=2147483647;
For X:=1 to 58 do {Find nearest pasture not
visited yet}
If Not Vis[X] and (Dist[X]<D) then
Begin
P:=X;
D:=Dist[X];
End;
If (P<>0) then
Begin
Vis[P]:=True; {If there is one mark it
visited}
For X:=1 to 58 do {And update all distances}
If (Conn[P,X]<>0) and (Dist[X]>Dist[P]+Conn[P,X]) then
Dist[X]:=Dist[P]+Conn[P,X];
End;
Until (P=0); {Until no reachable and unvisited pastures
left}
End;
Procedure Save;
Var TF :Text;
X,BD:LongInt; {BD = best distance}
BP :Char; {BP = best pasture}
Begin
BD:=2147483647;
For X:=1 to 25 do {Find neares pasture}
If (Dist[X]<BD) then
Begin
BD:=Dist[X];
BP:=Chr(Ord('A')+X-1);
End;
Assign(TF,'comehome.out');
Rewrite(TF);
Writeln(TF,BP,' ',BD); {Write outcome to disk}
Close(TF);
End;
Begin
Load;
Solve;
Save;
End.
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