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题解：一个超级神奇的题目，对于此题，首先很快可以想到用树状数组来搞，然后第一反应就是二维的，但想想这数据规模显然不现实。。。于是就想办法变成一维的，将Y坐标全部离散化（包括询问里面的），然后按照X坐标优先，Y坐标其次优先从小到大排序，然后遇到查询点就查询，否则就加入数组，然后没了，就是程序略长
1 /************************************************************** 2 Problem: 1935 3 User: HansBug 4 Language: Pascal 5 Result: Accepted 6 Time:7968 ms 7 Memory:100224 kb 8 ****************************************************************/ 9 10 var 11 i,j,k,l,m,n,nx,ny:longint; 12 a:array[0..600000,1..2] of longint; 13 b:array[0..1200000,1..2] of longint; 14 c:array[0..4000000] of longint; 15 d:array[0..3000000,1..2] of longint; 16 e:array[0..3000000,1..4] of longint; 17 procedure swap(var x,y:longint); 18 var z:longint; 19 begin 20 z:=x;x:=y;y:=z; 21 end; 22 procedure sort(l,r:longint); 23 var i,j,x,y:longint; 24 begin 25 i:=l;j:=r;x:=d[(l+r) div 2,1]; 26 repeat 27 while d[i,1]<x do inc(i); 28 while d[j,1]>x do dec(j); 29 if i<=j then 30 begin 31 swap(d[i,1],d[j,1]); 32 swap(d[i,2],d[j,2]); 33 inc(i);dec(j); 34 end; 35 until i>j; 36 if i<r then sort(i,r); 37 if l<j then sort(l,j); 38 end; 39 procedure sort0(l,r:longint); 40 var i,j,x,y,z:longint; 41 begin 42 i:=l;j:=r;x:=e[(l+r) div 2,1];y:=e[(l+r) div 2,2];z:=e[(l+r) div 2,3]; 43 repeat 44 while (e[i,1]<x) or ((e[i,1]=x) and (e[i,2]<y)) or ((e[i,1]=x) and (e[i,2]=y) and (e[i,3]>z)) do inc(i); 45 while (e[j,1]>x) or ((e[j,1]=x) and (e[j,2]>y)) or ((e[j,1]=x) and (e[j,2]=y) and (e[j,3]<z)) do dec(j); 46 if i<=j then 47 begin 48 swap(e[i,1],e[j,1]); 49 swap(e[i,2],e[j,2]); 50 swap(e[i,3],e[j,3]); 51 inc(i);dec(j); 52 end; 53 until i>j; 54 if i<r then sort0(i,r); 55 if l<j then sort0(l,j); 56 end; 57 procedure sort1(l,r:longint); 58 var i,j,x,y:longint; 59 begin 60 i:=l;j:=r;x:=e[(l+r) div 2,3]; 61 repeat 62 while e[i,3]<x do inc(i); 63 while e[j,3]>x do dec(j); 64 if i<=j then 65 begin 66 swap(e[i,1],e[j,1]); 67 swap(e[i,2],e[j,2]); 68 swap(e[i,3],e[j,3]); 69 swap(e[i,4],e[j,4]); 70 inc(i);dec(j); 71 end; 72 until i>j; 73 if i<r then sort1(i,r); 74 if l<j then sort1(l,j); 75 end; 76 procedure add(x,y:longint); 77 begin 78 if x<=0 then exit; 79 while x<=nx do 80 begin 81 inc(c[x],y); 82 inc(x,x and (-x)); 83 end; 84 end; 85 function sum(x:longint):longint; 86 begin 87 sum:=0; 88 while x>0 do 89 begin 90 inc(sum,c[x]); 91 dec(x,x and (-x)); 92 end; 93 end; 94 begin 95 readln(n,m); 96 for i:=1 to n do readln(a[i,1],a[i,2]); 97 for i:=1 to m do readln(b[i*2-1,1],b[i*2-1,2],b[i*2,1],b[i*2,2]); 98 for i:=1 to n do 99 begin 100 d[i,1]:=a[i,2]; 101 d[i,2]:=i; 102 end; 103 for i:=1 to m*2 do 104 begin 105 d[n+i,1]:=b[i,2]; 106 d[n+i,2]:=-i; 107 end; 108 sort(1,n+m*2);j:=0;d[0,1]:=-1; 109 for i:=1 to n+m*2 do 110 begin 111 if d[i,1]<>d[i-1,1] then inc(j); 112 if d[i,2]>0 then a[d[i,2],2]:=j else b[-d[i,2],2]:=j; 113 end; 114 nx:=j; 115 for i:=1 to m do 116 begin 117 e[i*4-3,1]:=b[i*2-1,1]-1;e[i*4-3,2]:=b[i*2-1,2]-1; 118 e[i*4-2,1]:=b[i*2-1,1]-1;e[i*4-2,2]:=b[i*2,2]; 119 e[i*4-1,1]:=b[i*2,1];e[i*4-1,2]:=b[i*2-1,2]-1; 120 e[i*4,1]:=b[i*2,1];e[i*4,2]:=b[i*2,2]; 121 end; 122 for i:=1 to n do begin e[m*4+i,1]:=a[i,1];e[m*4+i,2]:=a[i,2]; end; 123 for i:=1 to m*4+n do e[i,3]:=i; 124 sort0(1,m*4+n); 125 fillchar(c,sizeof(c),0); 126 for i:=1 to m*4+n do if e[i,3]<=(m*4) then e[i,4]:=sum(e[i,2]) else add(e[i,2],1); 127 sort1(1,m*4+n); 128 for i:=1 to m do writeln(e[i*4,4]-e[i*4-1,4]-e[i*4-2,4]+e[i*4-3,4]); 129 readln; 130 end.
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原文地址：http://www.cnblogs.com/HansBug/p/4492284.html
