POJ1683 Puzzlestan ——Floyd传递闭包+Dfs
好久没写Dfs了，拿来练手。
WA了一次，没有判断中间的情况……
解法：先用Floyd传递闭包处理哪些点一定要在一起、哪些点一定不能在一起，六重循环。
然后深搜，res[i][j]表示1,i这个物品在j这一行的匹配物品列编号。
没有最优性剪枝，只有一堆可能性剪枝：
(1)对于和(1,i)这个点关系为“一定在一起”的点(j,k)，一定要将res[i,j]设置为k。
(2)即将搜索res[i][j]为k的情况是否可能，那么条件就要是(j,k)这个点与(1,i)点的关系不是“不可能在一起”，而且(j,k)这个点与所有已经和(1,i)点匹配的点的关系不是“不可能在一起”。
参考代码：
program poj1683;//By_Thispoet
const maxn=10;
var
x1,y1,x2,y2 :longint;
i,j,m,n,p,q,test :longint;
map :array[0..maxn,0..maxn,0..maxn,0..maxn]of integer;
res :array[0..maxn,0..maxn]of integer;
v :array[0..maxn,0..maxn]of boolean;
ch :array[0..maxn,0..maxn]of char;
c,cc :char;
flag :boolean;
procedure printf();
begin
for i:=1 to m do begin
for j:=1 to n do write(ch[j,res[i,j]]);
writeln;
end;
writeln;
end;
procedure dfs(code,pos:longint);var i,j,k:longint;begin
if code=n+1 then begin
printf();flag:=true;exit;
end;
if res[pos,code]<>-1 then begin
if pos+1<=m then dfs(code,pos+1) else dfs(code+1,1);exit;
end;
if flag then exit;
for i:=1 to m do if (not v[code][i])and(map[1,pos,code,i]<>2) then begin
for j:=code+1 to n do for k:=1 to m do if (map[code,i,j,k]=1)and(res[pos,j]<>-1)then begin
flag:=true;break;
end;
for j:=1 to code-1 do if map[code,i,j,res[pos,j]]=2 then begin
flag:=true;break;
end;
if flag then begin flag:=false; continue; end;
v[code][i]:=true;res[pos,code]:=i;
for j:=code+1 to n do for k:=1 to m do if (map[code,i,j,k]=1) then begin
res[pos,j]:=k;v[j,k]:=true;break;
end;
if pos+1<=m then dfs(code,pos+1) else dfs(code+1,1);
if flag then exit;
for j:=code+1 to n do for k:=1 to m do if map[code,i,j,k]=1 then begin
res[pos,j]:=-1;v[j,k]:=false;break;
end;
v[code][i]:=false;res[pos,code]:=-1;
end;
end;
begin
readln(test);
while test>0 do begin
readln(n,m);filldword(map,sizeof(map)shr 2,0);
fillchar(res,sizeof(res),255);
for i:=1 to n do begin
for j:=1 to m do read(ch[i][j]);readln;
end;
flag:=false;readln(x1,y1,cc,c,cc,x2,y2);
while not (x1=0) do begin
if c='R' then begin
map[x1,y1,x2,y2]:=1;map[x2,y2,x1,y1]:=1;
end else begin
map[x1,y1,x2,y2]:=2;map[x2,y2,x1,y1]:=2;
end;
readln(x1,y1,cc,c,cc,x2,y2);
end;
for p:=1 to n do for q:=1 to m do
for x1:=1 to n do for y1:=1 to m do
for x2:=1 to n do for y2:=1 to m do
begin
if (map[x1,y1,p,q]=1)and(map[x2,y2,p,q]=2) then map[x1,y1,x2,y2]:=2;
if (map[x1,y1,p,q]=2)and(map[x2,y2,p,q]=1) then map[x1,y1,x2,y2]:=2;
if (map[x1,y1,p,q]=1)and(map[x2,y2,p,q]=1) then map[x1,y1,x2,y2]:=1;
end;
fillchar(v,sizeof(v),0);for i:=1 to m do res[i,1]:=i;
for i:=1 to m do for p:=2 to n do for q:=1 to m do
if map[1,i,p,q]=1 then begin res[i,p]:=q; v[p,q]:=true; end;
dfs(1,1);
dec(test);
end;
end.