[BZOJ3132] 上帝造题的七分钟

来源：互联网发布：巨衫数据库编辑：程序博客网时间：2024/04/20 02:34

传送门

http://www.lydsy.com/JudgeOnline/problem.php?id=3132

题目大意

支持矩阵加减+矩阵和查询

题解

二维树状数组

const    maxn=2048;type    data=array[0..maxn,0..maxn]of longint;var    c,d,e,f:data;    i,j,k:longint;    n,m,x1,x2,y1,y2,val,ans:longint;    cha:char;function query(var a:data;i,j:longint):longint;var sum,tt:longint;begin    sum:=0; tt:=j;    while i>0 do        begin            j:=tt;            while j>0 do                begin                    inc(sum,a[i,j]);                    dec(j,j and (-j));                end;            dec(i,i and (-i));        end;    exit(sum);end;function getsum(x,y:longint):longint;var sum:longint;begin    sum:=(x+1)*(y+1)*query(c,x,y)-(x+1)*query(e,x,y)-(y+1)*query(d,x,y)+query(f,x,y);    exit(sum);end;procedure update(var a:data;i,j,val:longint);var tt:longint;begin    tt:=j;    while i<=n do        begin            j:=tt;            while j<=m do                begin                    inc(a[i,j],val);                    inc(j,j and (-j));                end;            inc(i,i and (-i));        end;end;begin    readln(cha,n,m);    fillchar(c,sizeof(c),0);    fillchar(d,sizeof(d),0);    fillchar(e,sizeof(e),0);    fillchar(f,sizeof(f),0);    while not eof do        begin            read(cha);            if cha='k'            then                begin                    readln(x1,y1,x2,y2);                    ans:=getsum(x2,y2)-getsum(x1-1,y2)-getsum(x2,y1-1)+getsum(x1-1,y1-1);                    writeln(ans);                end            else                begin                    readln(x1,y1,x2,y2,val);                    update(c,x1,y1,val); update(c,x2+1,y1,-val); update(c,x1,y2+1,-val); update(c,x2+1,y2+1,val);                    update(d,x1,y1,val*x1); update(d,x2+1,y1,-val*(x2+1)); update(d,x1,y2+1,-val*x1); update(d,x2+1,y2+1,val*(x2+1));                    update(e,x1,y1,val*y1); update(e,x2+1,y1,-val*y1); update(e,x1,y2+1,-val*(y2+1)); update(e,x2+1,y2+1,val*(y2+1));                    update(f,x1,y1,val*x1*y1); update(f,x2+1,y1,-val*(x2+1)*y1); update(f,x1,y2+1,-val*x1*(y2+1)); update(f,x2+1,y2+1,val*(x2+1)*(y2+1));                end;        end;end.

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