bzoj2527

首先环可以变成链来处理，对于l>r的情况就是修改区间[1,r],[l,mx]
然后不难想到整体二分，二分答案k,然后算1~k场流星雨对国家的贡献
然后判定将国家划分变成子问题解决，没什么难的
终于不是tle，poi良心了一把

  1 type way=record
  2        po,next:longint;
  3      end;
  4      que=record
  5        p,n:longint;
  6      end;
  7      an=record
  8        l,r,v:longint;
  9      end;
 10 var a:array[0..300010] of an;
 11     qq,q:array[0..300010] of que;
 12     e:array[0..300010] of way;
 13     c:array[0..300010] of int64;
 14     p,ans,h:array[0..300010] of longint;
 15     v:array[0..300010] of boolean;
 16     tot,t,j,n,m,x,i:longint;
 17     s:int64;
 18 
 19 function lowbit(x:longint):longint;
 20   begin
 21     exit(x and (-x));
 22   end;
 23 
 24 procedure add(x,y:longint);
 25   begin
 26     e[i].po:=i;
 27     e[i].next:=p[x];
 28     p[x]:=i;
 29   end;
 30 
 31 procedure ins(x:longint;w:int64);
 32   begin
 33     while x<=n do
 34     begin
 35       if not v[x] then  //清理标记
 36       begin
 37         inc(tot);
 38         h[tot]:=x;
 39         v[x]:=true;
 40       end;
 41       c[x]:=c[x]+w;
 42       x:=x+lowbit(x);
 43     end;
 44   end;
 45 
 46 function ask(x:longint):int64;
 47   begin
 48     ask:=0;
 49     while x>0 do
 50     begin
 51       ask:=ask+c[x];
 52       x:=x-lowbit(x);
 53     end;
 54   end;
 55 
 56 procedure work(f,t,l,r:longint);
 57   var mid,l1,l2:longint;
 58   begin
 59     if f>t then exit;
 60     if l>r then exit;
 61     mid:=(l+r) shr 1;
 62     tot:=0;
 63     for i:=l to mid do
 64       if a[i].l<=a[i].r then
 65       begin
 66         ins(a[i].l,a[i].v);
 67         ins(a[i].r+1,-a[i].v);
 68       end
 69       else begin
 70         ins(1,a[i].v);
 71         ins(a[i].r+1,-a[i].v);
 72         ins(a[i].l,a[i].v);
 73       end;
 74 
 75     l1:=f;
 76     l2:=t;
 77     for i:=f to t do
 78     begin
 79       j:=p[q[i].p];
 80       s:=0;
 81       while j<>0 do
 82       begin
 83         s:=s+ask(e[j].po);
 84         if s>=q[i].n then
 85         begin
 86           qq[l1]:=q[i];
 87           inc(l1);
 88           ans[q[i].p]:=mid;
 89           break;
 90         end;
 91         j:=e[j].next;
 92       end;
 93       if s<q[i].n then
 94       begin
 95         q[i].n:=q[i].n-s;  //对于还不够的国家，直接把这部分贡献减去即可，下次直接处理mid之后的流星雨的贡献
 96         qq[l2]:=q[i];
 97         dec(l2);
 98       end;
 99     end;
100     for i:=1 to tot do
101     begin
102       c[h[i]]:=0;
103       v[h[i]]:=false;
104     end;
105     for i:=f to t do
106       q[i]:=qq[i];
107     work(f,l1-1,l,mid-1);
108     work(l2+1,t,mid+1,r);
109   end;
110 
111 begin
112   readln(m,n);
113   for i:=1 to n do
114   begin
115     read(x);
116     add(x,i);
117   end;
118   for i:=1 to m do
119   begin
120     read(q[i].n);
121     q[i].p:=i;
122   end;
123   readln(t);
124   for i:=1 to t do
125     readln(a[i].l,a[i].r,a[i].v);
126   work(1,m,1,t);
127   for i:=1 to m do
128     if ans[i]=0 then writeln('NIE')
129     else writeln(ans[i]);
130 end.

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posted on 2015-04-06 21:56 acphile 阅读(209) 评论(0) 编辑收藏举报