大水题一只，开100个二维树状数组表示，每一种颜色就可以了，查询的复杂度只有logm*logn，非常快。由于最近手残的严重，我还是喜闻乐见的WA了一次。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 | program p1452; var x:array[1..305,1..305,1..105] of longint; f:array[1..305,1..305] of longint; i,j,n,m,q,k1,k2,k3,k4,k5,k6:longint; function lowbit(k:longint):longint; begin exit(k and (-k)); end; procedure add(k1,k2,k3,k4:longint); var i,j:longint; begin i:=k1; while i<=n do begin j:=k2; while j<=m do begin inc(x[i,j,k3],k4); j:=j+lowbit(j); end; i:=i+lowbit(i); end; end; function find(k1,k2,k3:longint):longint; var i,j,num:longint; begin i:=k1; num:=0; while i>0 do begin j:=k2; while j>0 do begin inc(num,x[i,j,k3]); j:=j-lowbit(j); end; i:=i-lowbit(i); end; exit(num); end; begin readln(n,m); fillchar(x,sizeof(x),0); for i:=1 to n do for j:=1 to m do begin read(f[i,j]); add(i,j,f[i,j],1); end; read(q); for i:=1 to q do begin read(k1); if k1=1 then begin readln(k2,k3,k4); add(k2,k3,f[k2,k3],-1); add(k2,k3,k4,1); f[k2,k3]:=k4; end else begin readln(k2,k4,k3,k5,k6); writeln(find(k4,k5,k6)+find(k2-1,k3-1,k6)-find(k2-1,k5,k6)-find(k4,k3-1,k6)); end; end; end. |
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