1671: [Usaco2005 Dec]Knights of Ni 骑士
Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 254 Solved: 163
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Description
Bessie is in Camelot and has encountered a sticky situation: she needs to pass through the forest that is guarded by the Knights of Ni. In order to pass through safely, the Knights have demanded that she bring them a single shrubbery. Time is of the essence, and Bessie must find and bring them a shrubbery as quickly as possible. Bessie has a map of of the forest, which is partitioned into a square grid arrayed in the usual manner, with axes parallel to the X and Y axes. The map is W x H units in size (1 <= W <= 1000; 1 <= H <= 1000). The map shows where Bessie starts her quest, the single square where the Knights of Ni are, and the locations of all the shrubberies of the land. It also shows which areas of the map can be traverse (some grid blocks are impassable because of swamps, cliffs, and killer rabbits). Bessie can not pass through the Knights of Ni square without a shrubbery. In order to make sure that she follows the map correctly, Bessie can only move in four directions: North, East, South, or West (i.e., NOT diagonally). She requires one day to complete a traversal from one grid block to a neighboring grid block. It is guaranteed that Bessie will be able to obtain a shrubbery and then deliver it to the Knights of Ni. Determine the quickest way for her to do so.
Input
Output
Sample Input
4 1 0 0 0 0 1 0
0 0 0 1 0 1 0 0
0 2 1 1 3 0 4 0
0 0 0 4 1 1 1 0
INPUT DETAILS:
Width=8, height=4. Bessie starts on the third row, only a few squares away
from the Knights.
Sample Output
HINT
这片森林的长为8，宽为4．贝茜的起始位置在第3行，离骑士们不远．
贝茜可以按这样的路线完成骑士的任务：北，西，北，南，东，东，北，东，东，南，南．她在森林的西北角得到一株她需要的灌木，然后绕过障碍把它交给在东南方的骑士．
Source
题解：又是一道灌水法。。。。usaco灌水题真多= =
不过幸亏骑士的个数貌似就1个。。。要是骑士再一大堆的话那就惨了TT
还是灌水法不解释
const dir:array[..,..] of longint=((,),(-,),(,),(,-));
type arr=array[..] of longint;
var
i,j,k,l,m,n,x0,x1,y0,y1,head,tot,ans:longint;
a,b,h:array[..,..] of longint;
d:array[..,..] of longint;
c,e:arr;
function min(x,y:longint):longint;
begin
if x<y then min:=x else min:=y;
end;
procedure bfs(x,y:longint;var c:arr);
var f,r,xx,yy,i:longint;
begin
f:=;r:=;d[,]:=x;d[,]:=y;d[,]:=;b[x,y]:=;
while f<r do
begin
for i:= to do
begin
xx:=d[f,]+dir[i,];
yy:=d[f,]+dir[i,];
if b[xx,yy]= then continue;
b[xx,yy]:=;
d[r,]:=d[f,]+;
d[r,]:=xx;d[r,]:=yy;
if h[xx,yy]> then
begin
c[h[xx,yy]]:=d[r,];
end;
inc(r);
end;
inc(f);
end;
end;
begin
randomize;tot:=;
readln(m,n);
for i:= to n+ do
begin
a[i,m+]:=;
a[i,]:=;
end;
for i:= to m+ do
begin
a[n+,i]:=;
a[,i]:=;
end;
fillchar(h,sizeof(h),);
for i:= to n do
for j:= to m do
begin
read(a[i,j]);
case a[i,j] of
:begin
x0:=i;y0:=j;
end;
:begin
x1:=i;y1:=j;
a[i,j]:=;
end;
:begin
inc(tot);
h[i,j]:=tot;
a[i,j]:=;
end;
end;
if ((j mod )=) or (j=m) then readln;
end;
for i:= to n+ do
for j:= to m+ do
b[i,j]:=a[i,j];
fillchar(c,sizeof(c),);
bfs(x0,y0,c);
for i:= to n+ do
for j:= to m+ do
b[i,j]:=a[i,j];
fillchar(e,sizeof(e),);
bfs(x1,y1,e);
ans:=maxlongint;
for i:= to tot do
begin
if (c[i]=) or (e[i]=) then continue;
ans:=min(ans,(e[i]+c[i]))
end;
writeln(ans);
readln;
end.