POJ2479

题意:求两段不相交的连续子序列的和的最大值.如下:

水题,以前刚学线段树的时候写的,想到可以用线段树,一激动写了NlogN的.其实可以O(N)的.

code(600多MS,勿看):

type  node=record
      l,r,num:longint;
end;
const oo=600000000;
      maxn=51000;

var   f1,f2,a:array[0..51000] of longint;
      t:array[0..maxn*5] of node;
      datanum,d,n,s,ans,i,max1,max2:longint;

      function max(a,b:longint):longint;
      begin
            if a>b then exit(a); exit(b);
      end;

      procedure build(l,r,i:longint);
      var   mid:longint;
      begin
            t[i].l:=l;
            t[i].r:=r;
            if l=r then
            begin t[i].num:=f2[l]; exit; end;
            mid:=(l+r) shr 1;
            build(l,mid,i<<1);
            build(mid+1,r,i<<1+1);
            t[i].num:=max(t[i<<1].num,t[i<<1+1].num);
      end;

      procedure change(k,num,i:longint);
      begin
            if (t[i].r=t[i].l) then
              begin t[i].num:=num; exit; end
            else begin
                       if (k<=t[i<<1].r) then change(k,num,i<<1)
                         else change(k,num,i<<1+1);
                       t[i].num:=max(t[i<<1].num,t[i<<1+1].num);
                 end;
      end;

      function getans(l,r,i:longint):longint;
      var   ans1,ans2,mid:longint;
      begin
            if (t[i].l=l)and(t[i].r=r) then exit(t[i].num);
            ans1:=-oo;
            ans2:=-oo;
            mid:=(t[i].l+t[i].r)>>1;
            if r<=mid then ans1:=getans(l,r,i<<1)
            else if l>mid then ans2:=getans(l,r,i<<1+1)
                 else begin
                            ans1:=getans(l,mid,i<<1);
                            ans2:=getans(mid+1,r,i<<1+1);
                      end;
            exit(max(ans1,ans2));
      end;
begin
      readln(datanum);
      readln;
      for d:=1 to datanum do
      begin
            readln(n);
            fillchar(f1,sizeof(f1),0);
            fillchar(f2,sizeof(f2),0);
            for i:=1 to n do read(a[i]);

            for i:=1 to n do
            f1[i]:=max(f1[i-1]+a[i],a[i]);

            for i:=n downto 1 do
            f2[i]:=max(f2[i+1]+a[i],a[i]);

            build(1,n,1);
            ans:=-oo;
            for i:=1 to n-1 do
            begin
                  max1:=f1[i];
                  max2:=getans(i+1,n,1);
                  if max1+max2>ans then
                  ans:=max1+max2;
            end;
            writeln(ans);
            readln;
      end;
end.