POJ 3026 [Borg Maze]

　　这个题应该先求出每两个点对的最短距离（BFS），然后求最小生成树（Prim），表示BFS写得很疵，没有很好地注意边界条件

虽然AC了……

Type
        lzy=record
                x,y:longint;
end;
        Kudo=record
                pos,step,x,y:longint;
end;
Const
        dx:array[1..4]of -1..1=(1,-1,0,0);
        dy:array[1..4]of -1..1=(0,0,-1,1);
Var
        map:array[-1..500,-1..500]of char;
        spot,dis:array[-1..500,-1..500]of longint;
        link:array[-1..50*500]of lzy;
lowcost:array[1..50*500]of longint;
        vis:array[-1..500,-1..500]of boolean;
        q:array[-1..50*50000]of kudo;
        m,n,p,i,j,fuck,fucks:longint;
Procedure init;
var
        i,j:longint;
        ch:char;
begin
        readln(m,n);
        p:=0;
        fillchar(spot,sizeof(spot),0);
        fillchar(dis,sizeof(dis),$3f);
for i:=1 to n do
begin
for j:=1 to m do
begin
                                        read(map[i,j]);
if (map[i,j]='A')or(map[i,j]='S') then
begin
                                                        inc(p);
                                                        spot[i,j]:=p;
                                                        link[p].x:=i;
                                                        link[p].y:=j;
end;
end;
                        readln;
end;
end;
Procedure refresh(x1,y1,x2,y2,tmp:longint);
var
        left,right:longint;
begin
        left:=spot[x1,y1];
        right:=spot[x2,y2];
if tmp<dis[left,right] then
begin
                        dis[left,right]:=tmp;
                        dis[right,left]:=tmp;
end;
end;
Procedure getdis(stx,sty:longint);
var
        head,tail,i,j:longint;
        pre:Kudo;
        tmp:lzy;
        ch:char;
begin
        head:=0;
        tail:=1;
        fillchar(vis,sizeof(vis),false);
        q[1].pos:=spot[stx,sty];q[1].x:=stx;q[1].y:=sty;q[1].step:=0;
        vis[stx,sty]:=true;
while head<tail do
begin
                        inc(head);
                        pre:=q[head];
for i:=1 to 4 do
begin
                                        tmp.x:=pre.x+dx[i];
                                        tmp.y:=pre.y+dy[i];
if (tmp.x>0)and(tmp.x<=n)and(tmp.y>0)and(tmp.y<=m) then
begin
                                        ch:=map[tmp.x,tmp.y];

if (ch<>'#')and not vis[tmp.x,tmp.y] then
begin
                                                        inc(tail);
                                                        q[tail].x:=tmp.x;
                                                        q[tail].y:=tmp.y;
                                                        vis[tmp.x,tmp.y]:=true;
                                                        q[tail].pos:=spot[tmp.x,tmp.y];
                                                        q[tail].step:=q[head].step+1;
if (ch='A')or(ch='S') then refresh(tmp.x,tmp.y,stx,sty,q[tail].step);
end;
end;
end;
if tail>2500 then exit;
end;
end;
Function prim:longint;
var
        i,j,k,dist,sum,min:longint;
begin
        sum:=0;
        lowcost[1]:=-1;
for i:=2 to p do lowcost[i]:=dis[1,i];
for i:=1 to p-1 do
begin
                        min:=maxlongint;
for k:=1 to p do
begin
if (lowcost[k]<>-1)and(lowcost[k]<min) then
begin
                                                        j:=k;
                                                        min:=lowcost[k];
end;
end;
                        sum:=sum+min;
                        lowcost[j]:=-1;
for k:=1 to p do if dis[j,k]<lowcost[k] then lowcost[k]:=dis[j,k];
end;
        exit(sum);
end;
Begin
        readln(fucks);
for fuck:=1 to fucks do
begin
                        init;
for i:=1 to p do getdis(link[i].x,link[i].y);
               writeln(prim);
end;
End.

转载于:https://www.cnblogs.com/FreeDestiny/archive/2011/10/14/2212212.html