【图论】【poj 3026】Borg Maze

问题

S在迷宫中找A，找到A之后就把它同化，也会帮着S找剩下的A。。就是这样。。给你的是一个字符组成的图……

分析

正常人会想到搜索……当然，但是只是搜索很难实现，仔细读题，我们发现，其实质是在求最小生成树！！！

那么用宽搜的方法计算每两点之间的距离，重新构图，prim即可。为了方便查找，开一个which数组。点的坐标做下标，对应第几个点是数组中的值。

数据特别恶心，读入长和宽之后会出现n多莫名其妙的空格，要单独读图的第一行！！！

code

program liukee;
type xz=record
     x:longint;
     y:longint;
     z:longint;
end;

var
  d:array[1..4,1..2] of integer=((1,0),(-1,0),(0,1),(0,-1));
  x1,y1:array[1..101] of longint;
  map:array[0..51,0..51] of char;
  which:array[0..51,0..51] of longint;
  a:array[1..101,1..101] of longint;
  mincost:array[1..101] of longint;
  zu,n,m,ans,tt,i:longint;

procedure init;
var
  i,j:longint;
  xxx:char;
  s:string;
begin
  tt:=0;
  readln(m,n);
  readln(s);
  for i:=2to n do
  begin
    for j:=1to m do
    begin
      read(map[i,j]);
          xxx:=map[i,j];
      if (map[i,j]='A') or (map[i,j]='S') then
      begin
        inc(tt);
        which[i,j]:=tt;
        x1[tt]:=i;
        y1[tt]:=j;
      end;
    end;
    readln;
  end;
  for i:=1to m do
  begin
    map[1,i]:=s[i];
    if (map[1,i]='A') or (map[1,i]='S') then
      begin
        inc(tt);
        which[i,j]:=tt;
        x1[tt]:=i;
        y1[tt]:=j;
      end;
  end;
  for i:=length(s) to m do
    map[1,i]:='';
end;

procedure doit;
var
  q:array[1..100000] of xz;
  v:array[1..50,1..50] of boolean;
  i,l,r,xx,yy,j:longint;
begin
  for i:=1to tt do
  begin
    fillchar(q,sizeof(q),0);
    fillchar(v,sizeof(v),0);
    l:=0;
    r:=1;
    q[1].x:=x1[i];
    q[1].y:=y1[i];
    q[1].z:=0;
        v[x1[i],y1[i]]:=true;
    while l<r do
    begin
      inc(l);
      for j:=1to4do
      begin
        xx:=q[l].x+d[j,1];
        yy:=q[l].y+d[j,2];
        if(xx>0)and(yy>0)and(xx<=n)and(yy<=m)and(map[xx,yy]<>'#')and(not v[xx,yy])then
        begin
            if which[xx,yy]<>0then
            begin
              a[i,which[xx,yy]]:=q[l].z+1;
             // a[which[xx,yy],i]:=q[l].z+1;
            end;
          inc(r);
          q[r].x:=xx;
          q[r].y:=yy;
          q[r].z:=q[l].z+1;
          v[xx,yy]:=true;
        end;
      end;
    end;
  end;
end;

procedure prim;
var
  i,j,k,min:longint;
begin
  for i:=1to tt do
    for j:=1to tt do
      if a[i,j]=0then a[i,j]:=maxlongint;
  mincost[1]:=0;
  for i:=2to tt do
    mincost[i]:=a[1,i];
  for i:=1to tt-1do
  begin
    min:=maxlongint;
    for j:=1to tt do
      if (mincost[j]<min)and(mincost[j]<>0)then
      begin
        min:=mincost[j];
        k:=j;
      end;
    inc(ans,min);
    mincost[k]:=0;
    for j:=1to tt do
      if mincost[j]>a[k,j] then mincost[j]:=a[k,j];
  end;
end;

begin
  readln(zu);
  for i:=1to zu do
  begin
    ans:=0;
    fillchar(a,sizeof(a),0);
    fillchar(which,sizeof(which),0);
    fillchar(mincost,sizeof(mincost),0);
    init;
    doit;
    prim;
    writeln(ans);
  end;
end.

反思

要仔分析题目，抽象成图论模型。不要被表面现象迷惑。