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最新推荐文章于 2018-04-12 07:38:00 发布

转载最新推荐文章于 2018-04-12 07:38:00 发布 · 101 阅读

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原文链接：http://www.cnblogs.com/exponent/archive/2011/08/13/2137421.html

本文介绍了一种使用树状数组优化动态规划(DP)的方法，解决了一个特定的问题：如何高效地找到满足条件sum[j]<=sum[i]的所有j对应的f[j]之和。通过离散化和树状数组的应用，将原始O(N^2)的时间复杂度降低到更优的状态。

f[i]=∑f[j] (sum[j]<=sum[i]).

sum为前缀和.

裸的DP是O(N^2)的,其实我们要找的j只需满足sum[j]<=sum[i],然后求和.

而树状数组天生就是干这个的.范围大,离散化一下就好了.

code:

/************************************************************** 
    Problem: 2274
    User: exponent 
    Language: Pascal 
    Result: Accepted 
    Time:140 ms 
    Memory:2568 kb 
****************************************************************/ 
  
const maxn=100001; 
      mo=1000000009; 
var   s,t,p,q,f,c:array[0..maxn] of longint; 
      n,i,num,pre,pos,rank:longint; 
  
  
      procedure swap(var a,b:longint); 
      var   temp:longint; 
      begin
            temp:=a; a:=b; b:=temp; 
      end; 
  
      procedure sort(l,r:longint); 
      var   i,j,mid:longint; 
      begin
            i:=l; j:=r; 
            mid:=s[(l+r)>>1]; 
            while i<=j do
            begin
                  while s[i]<mid do inc(i); 
                  while s[j]>mid do dec(j); 
                  if i<=j then
                  begin
                        swap(s[i],s[j]); 
                        swap(p[i],p[j]); 
                        inc(i); dec(j); 
                  end; 
            end; 
            if i<r then sort(i,r); 
            if j>l then sort(l,j); 
      end; 
  
      function lowbit(i:longint):longint; 
      begin
            lowbit:=i and (i xor (i-1)); 
      end; 
  
      procedure change(i,delta:longint); 
      begin
            while i<=maxn do
            begin
                  c[i]:=(c[i]+delta) mod mo; 
                  inc(i,lowbit(i)); 
            end; 
      end; 
  
      function getsum(i:longint):longint; 
      begin
            getsum:=0; 
            while i>0 do
            begin
                  getsum:=(getsum+c[i]) mod mo; 
                  dec(i,lowbit(i)); 
            end; 
      end; 
  
begin
      readln(n); 
      for i:=1 to n do
      begin
            readln(num); 
            s[i]:=s[i-1]+num; 
            p[i]:=i; 
            q[i]:=s[i]; 
      end; 
      sort(1,n); 
      pre:=s[1]; 
      t[p[1]]:=1; 
      rank:=1; 
      for i:=2 to n do
         if s[i]=pre then t[p[i]]:=rank 
         else
         begin
               pre:=s[i]; 
               inc(rank); 
               t[p[i]]:=rank; 
         end; 
  
      f[0]:=1; 
      change(1,1); 
      for i:=1 to n do
      if q[i]>=0 then
      begin
            f[i]:=getsum(t[i]); 
            change(t[i],f[i]); 
      end; 
      writeln(f[n]); 
end.

转载于:https://www.cnblogs.com/exponent/archive/2011/08/13/2137421.html