vj p1037题解

原题叙述

这个题怎么说呢，还很能唬人的....

写了个很丑的背包....
不过秒掉了！
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编译通过...
├ 测试数据 01：答案正确... 0ms
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├ 测试数据 06：答案正确... 0ms
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├ 测试数据 10：答案正确... 0ms
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Accepted 有效得分：100 有效耗时：0ms

题解：

f[i,j]表示前i个数搭建的塔的差为j时的最大高度

v[i,j]为boolean数组，前i个数能否到达j的高度

动态方程：

f[i,j]:=max(f[i-1,j-a[i]]+a[i],f[i,j]);(if v[i-1,j-a[i]])

f[i,j]:=max(f[i-1,j+a[i]],f[i,j]);(if v[i-1,j+a[i]])

f[i,j]:=max(f[i-1,j],f[i,j]);if v[i-1,j+a[i]]

代码如下： 

 1 var n,all,i,j:longint;
 2     t:boolean;
 3     s,a:array[0..100]of longint;
 4     f:array[boolean,-2100..2100]of longint;
 5     v:array[boolean,-2100..2100]of boolean;
 6 function max(a,b,c:longint):longint;
 7 begin
 8      max:=a;
 9      if b>max then max:=b;
10      if c>max then max:=c;
11 end;
12 begin
13     
14      readln(n);
15      all:=0;
16      s[0]:=0;
17      for i:=1 to n do
18          begin
19               read(a[i]);
20               s[i]:=s[i-1]+a[i];
21          end;
22      fillchar(f,sizeof(f),0);
23      fillchar(v,sizeof(v),0);
24      t:=true;
25      v[false,0]:=true;
26      v[true,0]:=true;
27      for i:=1 to n do
28          begin
29               for j:=-s[i] to s[i] do
30                   begin
31                        f[t,j]:=0;v[t,j]:=false;
32                        if v[not(t),j+a[i]] then
33                           begin
34                                f[t,j]:=f[not(t),j+a[i]];
35                                v[t,j]:=true;
36                           end;
37                        if (v[not(t),j-a[i]])and(f[not(t),j-a[i]]+a[i]>=f[t,j]) then
38                           begin
39                                f[t,j]:=f[not(t),j-a[i]]+a[i];
40                                v[t,j]:=true;
41                           end;
42                        if (v[not(t),j])and(f[not(t),j]>=f[t,j]) then
43                           begin
44                                f[t,j]:=f[not(t),j];
45                                v[t,j]:=true;
46                           end;
47 //                       write(j,':',f[t,j],' ');
48                   end;
49 //              writeln;
50               t:=not(t);
51          end;
52      if v[true,0] and v[false,0] then
53         begin
54              if f[true,0]>f[false,0] then
55                 begin
56                      if f[true,0]=0 then writeln('Impossible') else writeln(f[true,0]);
57                 end
58              else
59                  begin
60                       if f[false,0]=0 then writeln('Impossible')
61                       else writeln(f[false,0]);
62                  end;
63         end
64      else
65          if (v[true,0])and(f[true,0]<>0) then writeln(f[true,0])
66          else
67              if (v[false,0])and(f[true,0]<>0) then writeln(f[false,0])
68              else
69                  writeln('Impossible');
70 end.