poj2528 Mayor's posters 2011-12-20

Mayor's posters

Time Limit: 1000MSMemory Limit: 65536K

Total Submissions: 23344Accepted: 6747

Description

 

The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters and introduce the following rules: 

Every candidate can place exactly one poster on the wall. 

All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown). 

The wall is divided into segments and the width of each segment is one byte. 

Each poster must completely cover a contiguous number of wall segments.

 

They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections. 

Your task is to find the number of visible posters when all the posters are placed given the information about posters' size, their place and order of placement on the electoral wall. 

Input

 

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among the n lines contains two integer numbers li and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= li <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered li, li+1 ,... , ri.

Output

 

For each input data set print the number of visible posters after all the posters are placed. 

 

The picture below illustrates the case of the sample input. 

 

Sample Input

 

1

5

1 4

2 6

8 10

3 4

7 10

Sample Output

 

4

Source

 

Alberta Collegiate Programming Contest 2003.10.18

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算是线段树的经典题吧，很早以前用非递归线段树做的。

题目就是在一堵墙上按顺序贴海报，最后询问有多少张海报可以被看见。有多组数据。

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需要离散化，也没什么特别的了。

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program Stone;

const t2=1 shl 15-1;

var i,j,k,l,c,n,ans:longint;

    a:array[1..2*t2+2]of longint;

    p:array[1..10000,1..2]of longint;

    h,s:array[1..20000]of longint;

    b:array[1..10000]of boolean;

procedure kp(t,w:longint);

var i,j,k,mid:longint;

 begin

  i:=t;j:=W;mid:=h[(t+w)div 2];

  repeat

   while h[i]<mid do inc(i);

   while h[j]>mid do dec(j);

    if i<=j then begin

                  k:=h[i];h[i]:=h[j];h[j]:=k;

                  inc(i);dec(j);

                 end;

  until i>j;

  if i<w then kp(i,w);

  if j>t then kp(t,j);

 end;

function find(x:longint):longint;  

var i,t,w:longint;

 begin

  t:=1;w:=2*n;

  repeat

    i:=(t+w)div 2;

    if h[i]<x then t:=i+1;

    if h[i]>x then w:=i-1;

    if h[i]=x then begin t:=i;break;end;

  until t>=w;

  find:=s[t];

 end; 

procedure add(x,y,z:longint);   //修改

var i,j,k:longint;

 begin

  x:=x+t2-1;y:=y+t2+1;

  while (x xor y)<>1 do

   begin

    if (x and 1)=0 then a[x+1]:=z;

    if (y and 1)=1 then a[y-1]:=z;

    x:=x div 2;y:=y div 2;

   end;

 end;

procedure init;

var i:longint;

 begin

  readln(n);

  for i:=1 to n do

   begin

    readln(p[i,1],p[i,2]);

    h[i*2-1]:=p[i,1];h[i*2]:=p[i,2];     //将所有海报左右坐标存成线性表。

   end;

  kp(1,2*n);j:=1;s[1]:=1;               

  for i:=2 to 2*n do

   begin

    if h[i]<>h[i-1] then inc(j);

    s[i]:=j;

   end;                                   //排序，离散化标号。

  for i:=1 to n do

   begin

    p[i,1]:=find(p[i,1]);p[i,2]:=find(p[i,2]);

    add(p[i,1]+1,p[i,2]+1,i);

   end;

 end;

procedure dfs(x,y:longint);

 begin

  if a[x]>y then y:=a[x];

  if (x>=t2+1) then begin

                    if (y<>0)and(b[y]) then begin

                                             inc(ans);b[y]:=false;

                                            end;

                    exit;

                   end;

  dfs(x*2,y);

  dfs(x*2+1,y);

 end;

begin

 assign(Input,'pku2528.in');assign(output,'pku2528.out');

 reset(input);rewrite(output);

  readln(c);

  for i:=1 to c do

   begin

    fillchar(a,sizeof(a),0);

    fillchar(h,sizeof(h),0);

    fillchar(s,sizeof(s),0);

    fillchar(b,sizeof(b),true);

    init;

    ans:=0;

    dfs(1,0);

    writeln(ans);

   end;

 close(input);close(output);

end.