zoj3956(Course Selection System)_Solution

zoj3956_Solution

H=sum(hi),C=sum(ci),Value=H*H-H*C-C*C

求Value的最大值

Solution：

动态规划：

共两维：H,C           固定一维C，在该维值C相同的情况下另一维应最大H，从而动态规划另一维H，转变为01背包问题。

优化：

H*H-H*C-C*C=0 (H,C>0)

H/C=(1+sqrt(5))/2=1.6180…

必会选择h/c>(1+sqrt(5))/2 的(h,c)对

证明：

若Value大于0，则H/C>(1+sqrt(5))/2，

若再加上h/c>(1+sqrt(5))/2 的(h,c)对，

(H+h)* (H+h)-(H+h) *(C+c)-(C+c)* (C+c)=H*H-H*C-C*C+h*h-h*c-c*c+2*H*h-H*c-h*C-2*C*c

[ (H+h)* (H+h)-(H+h) *(C+c)-(C+c)* (C+c) ] – [ H*H-H*C-C*C ]

= h*h-h*c-c*c+2*H*h-H*c-h*C-2*C*c

>2*H*h-H*c-h*C-2*C*c

>2*H*h - H*h*(1+sqrt(5))/2 - h*H*(1+sqrt(5))/2 – 2*H*(1+sqrt(5))/2*h*(1+sqrt(5))/2

=0

值必然会增大，所以必会选择h/c>(1+sqrt(5))/2 的(h,c)对，得证

所以可以把h/c>=(1+sqrt(5))/2的(h,c)对先选出来，再对h/c<(1+sqrt(5))/2的(h,c)对进行动态规划，然后在动态规划的基础上再加上h/c>=(1+sqrt(5))/2的(h,c)对求得在相同C的基础上H的最大值。

某个贪心的方法：按照h/c从大到小的顺序依次对h,c对进行排序，然后依次选取，直到值小于0。

证明不成立：

原来                 H,C                             value

加入                 h1,c1                         value1

再加入             h2,c2                         value2

（h1/c1>h2/c2）

若原来加入    h2,c2                         value3

证明有可能只加入h2,c2数值更大：

h1,c1数值非常大，且h1/c1<(1+sqrt(5))/2，加入h1,c1后value1<0，而再加入h2,c2后，value2<value1<0；而尽管h2/c2< h1/c1<(1+sqrt(5))/2，但加入h2,h2后，value2>value。

如H=1700,C=1000

h1=16000,c1=10000

h2=159,c1=100

value=190000

value1=-2410000

value2=-2501019

value3=200981

当有三个数据(1700,1000),(16000,10000),(159,100)，则用此方法求得结果为190000，并不正确。所以该方法错误。

DP：

 #include <stdio.h>
 #include <stdlib.h>
 #define maxn 500

 int main()
 {
     //500*100(total) *500(n) *10(test cases)=250000000

     //max_result=(10000*500)^2
     //the maximum of h[i] is larger than c[i],thus choose c
     //the total of c is not larger than 500*100=50000
     //when c is the same , we need maximum h
     long h[maxn+],c[maxn+];
     //<=10000/100*500
     long t,n,i,j,k,ans[maxn+],f[];
     long long s,result;
     scanf("%ld",&t);
     for (k=;k<=t;k++)
     {
         ans[]=;
         result=;
         scanf("%ld",&n);
         for (i=;i<=n;i++)
         {
             scanf("%ld %ld",&h[i],&c[i]);
             ans[i]=ans[i-]+c[i];
         }
         for (i=;i<=ans[n];i++)
             f[i]=-;
         f[]=;
         for (i=;i<=n;i++)
             //when choose course ith ,the maximum of credit is ans[i](including c[i])
             for (j=ans[i];j>=c[i];j--)
                 if (f[j-c[i]]!=- && f[j-c[i]]+h[i]>f[j])
                     f[j]=f[j-c[i]]+h[i];
         for (i=;i<=ans[n];i++)
             if (f[i]!=-)
             {
                 s=f[i]*f[i]-f[i]*i-i*i;
                 if (s>result)
                 result=s;
         }
         printf("%lld\n",result);
     }
     return ;
 }
 /*
 100 1 1 1 1 1 1 1 1
 0+101+102+……
 not obvious

 WorstSituation
 10 10 10 10 10
 50+50+50+50+50=250
 0+10+20+30+40=100
 save halt of the time

 Previous:
 49900*500=24950000
 Current:
 0+100+200+…+49900=12500000
 10Cases
 12500000*10=1,2500,0000
 */

DP(advance)：

 #include <stdio.h>
 #include <stdlib.h>
 #define maxn 500

 //原来60s，现在20s

 int main()
 {
     //500*100(total) *500(n) *10(test cases)=250000000

     //max_result=(10000*500)^2
     //the maximum of h[i] is larger than c[i],thus choose c
     //the total of c is not larger than 500*100=50000
     //when c is the same , we need maximum h
     long h[maxn+],c[maxn+],ans[maxn+],f[];
     //<=10000/100*500
     long t,n,i,j,k,g,temp,hh,cc;
     long long s,result;
     double line=(1.0+sqrt(5.0))/;
     scanf("%ld",&t);
     for (k=;k<=t;k++)
     {
         ans[]=;
         scanf("%ld",&n);
         for (i=;i<=n;i++)
             scanf("%ld %ld",&h[i],&c[i]);
         g=n+;
         for (i=n;i>=;i--)
             if (1.0*h[i]/c[i]>=line)
             {
                 g--;
                 temp=h[i];
                 h[i]=h[g];
                 h[g]=temp;
                 temp=c[i];
                 c[i]=c[g];
                 c[g]=temp;
             }
         g--;
         //a[g+1]~a[n] h[i]/c[i]>=line must be chose
         hh=;
         cc=;
         for (i=g+;i<=n;i++)
         {
             hh+=h[i];
             cc+=c[i];
         }
         //a[1]~a[g] h[i]/c[i]<line
         for (i=;i<=g;i++)
             ans[i]=ans[i-]+c[i];
         for (i=;i<=ans[g];i++)
             f[i]=-;
         f[]=;
         for (i=;i<=g;i++)
             //when choose course ith ,the maximum of credit is ans[i](including c[i])
             for (j=ans[i];j>=c[i];j--)
                 if (f[j-c[i]]!=- && f[j-c[i]]+h[i]>f[j])
                     f[j]=f[j-c[i]]+h[i];
         result=hh*hh-hh*cc-cc*cc;
         for (i=;i<=ans[g];i++)
             if (f[i]!=-)
             {
                 s=(f[i]+hh)*(f[i]+hh)-(f[i]+hh)*(i+cc)-(i+cc)*(i+cc);
                 if (s>result)
                 result=s;
         }
         printf("%lld\n",result);
     }
     return ;
 }
 /*
 100 1 1 1 1 1 1 1 1
 0+101+102+……
 not obvious

 WorstSituation
 10 10 10 10 10
 50+50+50+50+50=250
 0+10+20+30+40=100
 save halt of the time

 Previous:
 49900*500=24950000
 Current:
 0+100+200+…+49900=12500000
 10Cases
 12500000*10=1,2500,0000
 */

贪心(Wrong):

 #include <stdio.h>
 #include <stdlib.h>
 #define maxn 500

 struct node
 {
     long h,c;
     double per;
 };

 int cmp(const void *a,const void *b)
 {
     if ((*(struct node *)b).per>(*(struct node *)a).per)
         return ;
     else
         return ;
 }

 long long max(long long a,long long b)
 {
     if (a>b)
         return a;
     else
         return b;
 }

 int main()
 {
     long t,k,l,n,ht,ct;
     long long maxs;
     struct node course[maxn+];
     scanf("%ld",&t);
     for (k=;k<=t;k++)
     {
         scanf("%ld",&n);
         for (l=;l<n;l++)
         {
             scanf("%ld%ld",&course[l].h,&course[l].c);
             course[l].per=1.0*course[l].h/course[l].c;
         }
         qsort(course,n,sizeof(struct node),cmp);
         ht=;
         ct=;
         maxs=;
         for (l=;l<n;l++)
         {
             ht+=course[l].h;
             ct+=course[l].c;
             maxs=max(maxs,ht*ht-ht*ct-ct*ct);
         }
         printf("%lld\n",maxs);
     }
     return ;
 }
 /*
 5
 1 4
 2 5
 6 3
 7 2
 4 4
 */

最普通的方法，保证正确，用于测试其它程序是否正确：

 #include <stdio.h>
 #include <stdlib.h>

 struct node
 {
     long h,c;
     double per;
 };
     long n;
     long long maxs=;
         struct node course[];

 int cmp(const void *a,const void *b)
 {
     return (*(struct node *)b).per-(*(struct node *)a).per;
 }

 long long max(long long a,long long b)
 {
     if (a>b)
         return a;
     else
         return b;
 }

 void dfs(long d,long ht,long ct)
 {
     if (d==n)
         return ;
     if (ct!=)
         maxs=max(maxs,ht*ht-ht*ct-ct*ct);
     dfs(d+,ht+course[d].h,ct+course[d].c);
     dfs(d+,ht,ct);
 }

 int main()
 {
     long t,k,l;
     scanf("%ld",&t);
     for (k=;k<=t;k++)
     {
         scanf("%ld",&n);
         maxs=;
         for (l=;l<n;l++)
         {
             scanf("%ld%ld",&course[l].h,&course[l].c);
             course[l].per=1.0*course[l].h/course[l].c;
         }
         dfs(,,);
 printf("%lld\n",maxs);
     }
     return ;
 }

以下是我做的关于此题做的评测方法，值得一览

http://pan.baidu.com/s/1c1WeUGS