【CF676C】Vasya and String（二分查找，线性扫描尺取法）

题意：

给出一个长度为n的字符串，只有字符'a'和'b'。最多能改变k个字符，即把'a'变成'b'或把'b'变成'a'。

问改变后的最长连续相同字符的字串长度为多少。

首先是二分查找，好想也好写

var s:array[0..100000]of longint;
    ch:ansistring;
    n,k,i,l,r,mid,last,ans:longint;

function max(x,y:longint):longint;
begin
 if x>y then exit(x);
 exit(y);
end;

begin
 //assign(input,'1.in'); reset(input);
 //assign(output,'1.out'); rewrite(output);
 readln(n,k);
 readln(ch);
 for i:=1 to n do
 begin
  s[i]:=s[i-1];
  if ch[i]='b' then inc(s[i]);
 end;
 ans:=0;
 for i:=1 to n do
  begin
  l:=i; r:=n; last:=i;
  while l<=r do
  begin
   mid:=(l+r)>>1;
   if s[mid]-s[i-1]<=k then begin last:=mid; l:=mid+1; end
    else r:=mid-1;
  end;
  ans:=max(ans,last-i+1);
 end;
 fillchar(s,sizeof(s),0);
 for i:=1 to n do
 begin
  s[i]:=s[i-1];
  if ch[i]='a' then inc(s[i]);
 end;
 for i:=1 to n do
  begin
  l:=i; r:=n; last:=i;
  while l<=r do
  begin
   mid:=(l+r)>>1;
   if s[mid]-s[i-1]<=k then begin last:=mid; l:=mid+1; end
    else r:=mid-1;
  end;
  ans:=max(ans,last-i+1);
 end;
 writeln(ans);
 //close(input);
 //close(output);
end.

 然后是线性扫描，ACM叫做尺取法，机房大神叫伸头缩尾法

l循环后还要+1是因为要到下一段连续区间的开头，而当前连续间的字母和下一段一定不同（显然要找最长的连续相同序列），先默认将开头字母改好

var ch:ansistring;
    n,k,ans,r,l,i,t:longint;

function max(x,y:longint):longint;
begin
 if x>y then exit(x);
 exit(y);
end;

begin
 //assign(input,'1.in'); reset(input);
 //assign(output,'1.out'); rewrite(output);
 readln(n,k);
 readln(ch);
 l:=1; r:=1; t:=0;
 for i:=1 to n do
 begin
  if ch[i]='b' then
  begin
   if t<k then begin inc(t); inc(r); end
    else
    begin
     while (l<=n)and(ch[l]='a') do inc(l);
     inc(l);
     inc(r);
    end;
  end
   else inc(r);
  ans:=max(ans,r-l);
 end;
 l:=1; r:=1; t:=0;
 for i:=1 to n do
 begin
  if ch[i]='a' then
  begin
   if t<k then begin inc(t); inc(r); end
    else
    begin
     while (l<=n)and(ch[l]='b') do inc(l);
     inc(l);
     inc(r);
    end;
  end
   else inc(r);
  ans:=max(ans,r-l);
 end;
 writeln(ans);
 //close(input);
 //close(output);
end.