poj2528 Mayor's posters 2011-12-20

Posted 2020-06-17

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。

______________________________________________________

需要离散化，也没什么特别的了。

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  1 program Stone;
  2 
  3 const t2=1 shl 15-1;
  4 
  5 var i,j,k,l,c,n,ans:longint;
  6 
  7     a:array[1..2*t2+2]of longint;
  8 
  9     p:array[1..10000,1..2]of longint;
 10 
 11     h,s:array[1..20000]of longint;
 12 
 13     b:array[1..10000]of boolean;
 14 
 15 procedure kp(t,w:longint);
 16 
 17 var i,j,k,mid:longint;
 18 
 19  begin
 20 
 21   i:=t;j:=W;mid:=h[(t+w)div 2];
 22 
 23   repeat
 24 
 25    while h[i]<mid do inc(i);
 26 
 27    while h[j]>mid do dec(j);
 28 
 29     if i<=j then begin
 30 
 31                   k:=h[i];h[i]:=h[j];h[j]:=k;
 32 
 33                   inc(i);dec(j);
 34 
 35                  end;
 36 
 37   until i>j;
 38 
 39   if i<w then kp(i,w);
 40 
 41   if j>t then kp(t,j);
 42 
 43  end;
 44 
 45 function find(x:longint):longint;  
 46 
 47 var i,t,w:longint;
 48 
 49  begin
 50 
 51   t:=1;w:=2*n;
 52 
 53   repeat
 54 
 55     i:=(t+w)div 2;
 56 
 57     if h[i]<x then t:=i+1;
 58 
 59     if h[i]>x then w:=i-1;
 60 
 61     if h[i]=x then begin t:=i;break;end;
 62 
 63   until t>=w;
 64 
 65   find:=s[t];
 66 
 67  end; 
 68 
 69 procedure add(x,y,z:longint);   //修改
 70 
 71 var i,j,k:longint;
 72 
 73  begin
 74 
 75   x:=x+t2-1;y:=y+t2+1;
 76 
 77   while (x xor y)<>1 do
 78 
 79    begin
 80 
 81     if (x and 1)=0 then a[x+1]:=z;
 82 
 83     if (y and 1)=1 then a[y-1]:=z;
 84 
 85     x:=x div 2;y:=y div 2;
 86 
 87    end;
 88 
 89  end;
 90 
 91 procedure init;
 92 
 93 var i:longint;
 94 
 95  begin
 96 
 97   readln(n);
 98 
 99   for i:=1 to n do
100 
101    begin
102 
103     readln(p[i,1],p[i,2]);
104 
105     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。
106 
107    end;
108 
109   kp(1,2*n);j:=1;s[1]:=1;               
110 
111   for i:=2 to 2*n do
112 
113    begin
114 
115     if h[i]<>h[i-1] then inc(j);
116 
117     s[i]:=j;
118 
119    end;                                   //排序，离散化标号。
120 
121   for i:=1 to n do
122 
123    begin
124 
125     p[i,1]:=find(p[i,1]);p[i,2]:=find(p[i,2]);
126 
127     add(p[i,1]+1,p[i,2]+1,i);
128 
129    end;
130 
131  end;
132 
133 procedure dfs(x,y:longint);
134 
135  begin
136 
137   if a[x]>y then y:=a[x];
138 
139   if (x>=t2+1) then begin
140 
141                     if (y<>0)and(b[y]) then begin
142 
143                                              inc(ans);b[y]:=false;
144 
145                                             end;
146 
147                     exit;
148 
149                    end;
150 
151   dfs(x*2,y);
152 
153   dfs(x*2+1,y);
154 
155  end;
156 
157 begin
158 
159  assign(Input,‘pku2528.in‘);assign(output,‘pku2528.out‘);
160 
161  reset(input);rewrite(output);
162 
163   readln(c);
164 
165   for i:=1 to c do
166 
167    begin
168 
169     fillchar(a,sizeof(a),0);
170 
171     fillchar(h,sizeof(h),0);
172 
173     fillchar(s,sizeof(s),0);
174 
175     fillchar(b,sizeof(b),true);
176 
177     init;
178 
179     ans:=0;
180 
181     dfs(1,0);
182 
183     writeln(ans);
184 
185    end;
186 
187  close(input);close(output);
188 
189 end.