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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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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.
poj2528 Mayor's posters 2011-12-20
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原文地址：http://www.cnblogs.com/yesphet/p/5235827.html
