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RQNOJ-43·飞翔·乱解

人云E云 posted @ 2011年10月20日 03:17 in 信息技術 with tags 解题报告 RQNOJ , 1314 阅读

昨天有人问我RQNOJ的第43题，，，就是那道飞翔，当时没思路，晚上回家的时候想到了方案，今天早上给打了出来，A掉了……//其实我今天早上打的跟昨晚想的不太一样，我昨晚想的那个是个 k^3 的做法，，不过早上发现更优方案了

题目嘛，我感觉有些描述不清{我白痴嘛}……后来问了下，才明白从第三行起的东西是可以走对角线的正方形的左下角坐标

N×M，明显爆了。而k只有1000，这就给题弄出了一个切入点：把所有的特殊的正方形列出来，到某个正方形为止的所有走的对角线个数是该正方形左下角的所有正方形中走的对角线数的最大值。

在这里，可以直接将正方形当作点看，，以下“点”全是指这些特殊正方形

不过要注意，不要总是枚举它最近的点，这样是错误的，，反例自己就可以举出来。

于是，可以对所有的特殊点进行排序{对X或者对Y皆可}{我用X的小到大}，之后在循环中对这些点之前的点进行枚举，找出最大的个数x，那么经过该点之后的个数便是x+1

若设读取的点坐标记录为map[*][1] map[*][2]，走过的对角线个数记为num[*]，伪代码可以如下写

FOR i = 1 TO k {

FOR j = i-1 TO 0

IF map[j,1]<map[i,1] AND map[j,2]<map[i,2] THEN x = num[j]

num[i] = x+1

}

附上我的PASCAL代码……

var
 point:array[0..1000,1..2] of longint;//就是正文中的map
 line:array[0..1000] of longint;//就是正文中的num
 n,m,mm,i,j,x,y,z:longint;//mm就是题中的k
 Procedure change(i,j:longint);//交换数据
 Var
  x:longint;
 Begin
  x:=point[i,1]; point[i,1]:=point[j,1]; point[j,1]:=x;
  x:=point[j,2]; point[j,2]:=point[i,2]; point[i,2]:=x;
 End;
 Procedure qsort(l,r:longint);
 Var
  i,j,x,y:longint;
 Begin
  i:=l;
  j:=r;
  x:=point[(l+r) shr 1,1];
  y:=point[(l+r) shr 1,2];
   while i<=j do
    begin
      while (point[i,1]<x) or ((point[i,1]=x) and (point[i,2]<y)) do
       inc(i);
      while (point[j,1]>x) or ((point[j,1]=x) and (point[j,2]>y)) do
       dec(j);
      if i<=j then
       begin
        change(i,j);
        inc(i);
        dec(j);
        end;
     end;
   if i<r then
    qsort(i,r);
   if j>l then
    qsort(l,j);
 End;
begin
 readln(n,m);
 readln(mm);
 z:=0;
  for i:=1 to mm do
   begin
    readln(x,y);
    point[i,1]:=x;
    point[i,2]:=y;
    end;//这里我本来想离散化的，后来发现不需要{而且我也不会……}
 qsort1(1,mm);
 point[0,1]:=0;
 point[0,2]:=0;
 line[0]:=0;//赋一下初值
  for i:=1 to mm do
   begin
    z:=0;
     for j:=i-1 downto 0 do
      if (point[j,1]<point[i,1]) and (point[j,2]<point[i,2]) then
       if z<line[j] then
        z:=line[j];
    line[i]:=z+1;
    end;//见上文伪代码
 x:=0;
  for i:=1 to mm do
   if x<line[i] then
    x:=line[i];//求最大对角线个数
 z:=(n+m-x-x)*100+round(sqrt(20000)*x);
 writeln(z);
end.

后，，不会写伪代码……求教各位大牛……

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