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uoj6

2024-09-14 21:56:36   217人阅读

这道题看起来仿佛有点水

从小到大贪心，舍去左下角和右上角

空间卡得有点紧，用integer/short int

1、30分做法

     记录数字的顺序及每个数字的位置，然后对于每个数字的位置与前面已取数字的位置进行判断，时间复杂度（n3）

const maxnm=5000*5000+10;maxn=5000+10;
var pre,now,a,b,c,d,i1:int64;
    n,m,qu,nm,i,j,num,u,v,k:longint;
    t:array[0..maxnm] of longint;
    x:array[0..maxnm,1..2] of integer;
    bool:array[0..maxnm] of boolean;
    q:array[0..maxn*2] of longint;

procedure swap64(var x,y:int64);
var t:int64;
begin
  t:=x;x:=y;y:=t;
end;

procedure swaplong(var x,y:longint);
var t:longint;
begin
  t:=x;x:=y;y:=t;
end;

function check(x1,y1,x2,y2:longint):boolean;
begin
  if (x1<x2)and(y1>y2)or(x1>x2)and(y1<y2) then exit(false);
  exit(true);
end;

begin
read(pre,a,b,c,d);
read(n,m,qu);
nm:=n*m;
for i:=1 to nm do
  t[i]:=i;
for i:=1 to nm do
begin
  now:=(a*pre*pre+b*pre+c) mod d;
  i1:=i;
  swaplong(t[i],t[now mod i1+1]);
  swap64(now,pre);
end;
for i:=1 to qu do
begin
  read(u,v);
  swaplong(t[u],t[v]);
end;
num:=0;
for i:=1 to n do
  for j:=1 to m do
  begin
    inc(num);
    x[t[num],1]:=i;
    x[t[num],2]:=j;
  end;
num:=0;
for i:=1 to nm do bool[i]:=false;
for i:=1 to nm do
  for j:=1 to num+1 do
  if j=num+1 then
  begin
    bool[i]:=true;
    write(i,‘ ‘);
    inc(num);
    q[num]:=i;
  end
  else if check(x[i,1],x[i,2],x[q[j],1],x[q[j],2])=false then break;
end.

2、100分做法

     对于行，记录比自身大的行的最小列，比自身小的行的最大列，对每个数字进行判断，时间复杂度（n2）

     bzoj过，被UOJ卡到60。。。。。。

const maxnm=5000*5000+10;maxn=5000+10;oo=1000000000;
var pre,now,a,b,c,d,i1:int64;
    n,m,qu,nm,i,j,num,u,v:longint;
    t:array[0..maxnm] of longint;
    x:array[0..maxnm,1..2] of integer;
    line:array[0..maxn,0..1] of longint;

function min(x,y:longint):longint;
begin
  if x<y then min:=x else min:=y;
end;

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

procedure swap64(var x,y:int64);
var t:int64;
begin
  t:=x;x:=y;y:=t;
end;

procedure swaplong(var x,y:longint);
var t:longint;
begin
  t:=x;x:=y;y:=t;
end;

function check(x1,y1,x2,y2:longint):boolean;
begin
  if (x1<x2)and(y1>y2)or(x1>x2)and(y1<y2) then exit(false);
  exit(true);
end;

begin
read(pre,a,b,c,d);
read(n,m,qu);
nm:=n*m;
for i:=1 to nm do
  t[i]:=i;
for i:=1 to nm do
begin
  now:=(a*pre*pre+b*pre+c) mod d;
  i1:=i;
  swaplong(t[i],t[now mod i1+1]);
  swap64(now,pre);
end;
for i:=1 to qu do
begin
  read(u,v);
  swaplong(t[u],t[v]);
end;
num:=0;
for i:=1 to n do
  for j:=1 to m do
  begin
    inc(num);
    x[t[num],1]:=i;
    x[t[num],2]:=j;
  end;
for i:=1 to n do
begin
  line[i,0]:=0;
  line[i,1]:=oo;
end;
for i:=1 to nm do
  if not((line[x[i,1],0]>x[i,2])or(line[x[i,1],1]<x[i,2])) then
  begin
    write(i,‘ ‘);
    for j:=1 to x[i,1]-1 do line[j,1]:=min(line[j,1],x[i,2]);
    for j:=x[i,1]+1 to n do line[j,0]:=max(line[j,0],x[i,2]);
  end;
end.

　　

uoj6


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