bzoj2436

不难发现两边的活动是交替进行的，我们可以dp

先对时间离散化，设f[i,j]到时间i一个会场选j个活动，另一个会场最多有多少活动，那么f[i,j]=max(f[k,j]+s[k,i],f[k,j-s[k,i]])

然后第一问的答案就是max(min(f[n*2,i],i))

第二问是要求必选一个，那这一定形如中间一段必须选，向左向右分别dp

假设中间那段是(i,j)时最优值是g[i,j]

肯定要预处理时间i前一个会场选j个活动，另一个会场最多有多少活动和时间i到结束一个会场选j个活动，另一个会场最多有多少活动

我们记为fl[i,j],fr[i,j]，不难得到g[i,j]=max{min(x+y+s[i,j],fl[i,x]+fr[j,y])} （注意到f[]数组的两个会场取值是对称的）

处理g[i,j]是O(n4）的，拍几组数据可以发现当i,j固定时，随着x的增大，对应最优的y减小

其实主观上也很好理解，肯定是两个会场活动尽量平衡，证明吗……不会……

 var fl,fr,f,g:array[..,..] of longint;
     a,s,e:array[..] of longint;
     i,n,m,k,j,x,y,ans,t:longint;
 
 procedure max(var a:longint; b:longint);
   begin
     if a<b then a:=b;
   end;
 
 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;
 
 procedure swap(var a,b:longint);
   var c:longint;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;
 
 procedure sort(l,r:longint);
   var i,j,x:longint;
   begin
     i:=l;
     j:=r;
     x:=a[(l+r) shr ];
     repeat
       while a[i]<x do inc(i);
       while x<a[j] do dec(j);
       if not(i>j) then
       begin
         swap(a[i],a[j]);
         inc(i);
         dec(j);
       end;
     until i>j;
     if l<j then sort(l,j);
     if i<r then sort(i,r);
   end;
 
 function find(l,r,x:longint):longint;
   var m:longint;
   begin
     while l<=r do
     begin
       m:=(l+r) shr ;
       if a[m]=x then exit(m);
       if a[m]>x then r:=m- else l:=m+;
     end;
   end;
 
 begin
   readln(n);
   for i:= to n do
   begin
     readln(s[i],e[i]);
     e[i]:=s[i]+e[i];
     inc(t);
     a[t]:=s[i];
     inc(t);
     a[t]:=e[i];
   end;
   sort(,t);
   m:=;
   for i:= to t do
     if a[i]<>a[m] then
     begin
       inc(m);
       a[m]:=a[i];
     end;
   for i:= to n do
   begin
     s[i]:=find(,m,s[i]);
     e[i]:=find(,m,e[i]);
     inc(g[s[i],e[i]]);
   end;
   for i:= to m do
     for j:=i+ to m do
       inc(g[i,j],g[i,j-]);
   for j:= to m do
     for i:=j- downto  do
       inc(g[i,j],g[i+,j]);
   for i:= to m+ do
     for j:= to n do
     begin
       fl[i,j]:=-n;
       fr[i,j]:=-n;
     end;
   fl[,]:=; fr[m+,]:=;
   for i:= to m do
     for j:= to i- do
       for k:= to n do
       begin
         max(fl[i,k],fl[j,k]+g[j,i]);
         if k>=g[j,i] then max(fl[i,k],fl[j,k-g[j,i]]);
       end;
   for i:=m downto  do
     for j:=i+ to m+ do
       for k:= to n do
       begin
         max(fr[i,k],fr[j,k]+g[i,j]);
         if k>=g[i,j] then max(fr[i,k],fr[j,k-g[i,j]]);
       end;
   ans:=;
   for i:= to n do
     max(ans,min(fl[m,i],i));
   writeln(ans);
   for i:= to m do
     for j:=i to m do
     begin
       y:=g[j,m];
       for x:= to g[,i] do
       begin
         while (y>) and (min(x+y-+g[i,j],fl[i,x]+fr[j,y-])>=min(x+y+g[i,j],fl[i,x]+fr[j,y])) do dec(y);
         max(f[i,j],min(x+y+g[i,j],fl[i,x]+fr[j,y]));
       end;
     end;
 
   for i:= to n do
   begin
     ans:=;
     for x:= to s[i] do
       for y:=e[i] to m do
         max(ans,f[x,y]);
     writeln(ans);
   end;
 end.