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：
1 #include <stdio.h>
2 #include <stdlib.h>
3 #define maxn 500
4
5 int main()
6 {
7 //500*100(total) *500(n) *10(test cases)=250000000
8
9 //max_result=(10000*500)^2
10 //the maximum of h[i] is larger than c[i],thus choose c
11 //the total of c is not larger than 500*100=50000
12 //when c is the same , we need maximum h
13 long h[maxn+1],c[maxn+1];
14 //<=10000/100*500
15 long t,n,i,j,k,ans[maxn+1],f[50001];
16 long long s,result;
17 scanf("%ld",&t);
18 for (k=1;k<=t;k++)
19 {
20 ans[0]=0;
21 result=0;
22 scanf("%ld",&n);
23 for (i=1;i<=n;i++)
24 {
25 scanf("%ld %ld",&h[i],&c[i]);
26 ans[i]=ans[i-1]+c[i];
27 }
28 for (i=1;i<=ans[n];i++)
29 f[i]=-1;
30 f[0]=0;
31 for (i=1;i<=n;i++)
32 //when choose course ith ,the maximum of credit is ans[i](including c[i])
33 for (j=ans[i];j>=c[i];j--)
34 if (f[j-c[i]]!=-1 && f[j-c[i]]+h[i]>f[j])
35 f[j]=f[j-c[i]]+h[i];
36 for (i=1;i<=ans[n];i++)
37 if (f[i]!=-1)
38 {
39 s=f[i]*f[i]-f[i]*i-i*i;
40 if (s>result)
41 result=s;
42 }
43 printf("%lld\n",result);
44 }
45 return 0;
46 }
47 /*
48 100 1 1 1 1 1 1 1 1
49 0+101+102+……
50 not obvious
51
52 WorstSituation
53 10 10 10 10 10
54 50+50+50+50+50=250
55 0+10+20+30+40=100
56 save halt of the time
57
58 Previous:
59 49900*500=24950000
60 Current:
61 0+100+200+…+49900=12500000
62 10Cases
63 12500000*10=1,2500,0000
64 */
DP(advance)：
1 #include <stdio.h>
2 #include <stdlib.h>
3 #define maxn 500
4
5 //原来60s，现在20s
6
7 int main()
8 {
9 //500*100(total) *500(n) *10(test cases)=250000000
10
11 //max_result=(10000*500)^2
12 //the maximum of h[i] is larger than c[i],thus choose c
13 //the total of c is not larger than 500*100=50000
14 //when c is the same , we need maximum h
15 long h[maxn+1],c[maxn+1],ans[maxn+1],f[50001];
16 //<=10000/100*500
17 long t,n,i,j,k,g,temp,hh,cc;
18 long long s,result;
19 double line=(1.0+sqrt(5.0))/2;
20 scanf("%ld",&t);
21 for (k=1;k<=t;k++)
22 {
23 ans[0]=0;
24 scanf("%ld",&n);
25 for (i=1;i<=n;i++)
26 scanf("%ld %ld",&h[i],&c[i]);
27 g=n+1;
28 for (i=n;i>=1;i--)
29 if (1.0*h[i]/c[i]>=line)
30 {
31 g--;
32 temp=h[i];
33 h[i]=h[g];
34 h[g]=temp;
35 temp=c[i];
36 c[i]=c[g];
37 c[g]=temp;
38 }
39 g--;
40 //a[g+1]~a[n] h[i]/c[i]>=line must be chose
41 hh=0;
42 cc=0;
43 for (i=g+1;i<=n;i++)
44 {
45 hh+=h[i];
46 cc+=c[i];
47 }
48 //a[1]~a[g] h[i]/c[i]<line
49 for (i=1;i<=g;i++)
50 ans[i]=ans[i-1]+c[i];
51 for (i=1;i<=ans[g];i++)
52 f[i]=-1;
53 f[0]=0;
54 for (i=1;i<=g;i++)
55 //when choose course ith ,the maximum of credit is ans[i](including c[i])
56 for (j=ans[i];j>=c[i];j--)
57 if (f[j-c[i]]!=-1 && f[j-c[i]]+h[i]>f[j])
58 f[j]=f[j-c[i]]+h[i];
59 result=hh*hh-hh*cc-cc*cc;
60 for (i=1;i<=ans[g];i++)
61 if (f[i]!=-1)
62 {
63 s=(f[i]+hh)*(f[i]+hh)-(f[i]+hh)*(i+cc)-(i+cc)*(i+cc);
64 if (s>result)
65 result=s;
66 }
67 printf("%lld\n",result);
68 }
69 return 0;
70 }
71 /*
72 100 1 1 1 1 1 1 1 1
73 0+101+102+……
74 not obvious
75
76 WorstSituation
77 10 10 10 10 10
78 50+50+50+50+50=250
79 0+10+20+30+40=100
80 save halt of the time
81
82 Previous:
83 49900*500=24950000
84 Current:
85 0+100+200+…+49900=12500000
86 10Cases
87 12500000*10=1,2500,0000
88 */
贪心(Wrong):
1 #include <stdio.h>
2 #include <stdlib.h>
3 #define maxn 500
4
5 struct node
6 {
7 long h,c;
8 double per;
9 };
10
11 int cmp(const void *a,const void *b)
12 {
13 if ((*(struct node *)b).per>(*(struct node *)a).per)
14 return 1;
15 else
16 return 0;
17 }
18
19 long long max(long long a,long long b)
20 {
21 if (a>b)
22 return a;
23 else
24 return b;
25 }
26
27 int main()
28 {
29 long t,k,l,n,ht,ct;
30 long long maxs;
31 struct node course[maxn+1];
32 scanf("%ld",&t);
33 for (k=1;k<=t;k++)
34 {
35 scanf("%ld",&n);
36 for (l=0;l<n;l++)
37 {
38 scanf("%ld%ld",&course[l].h,&course[l].c);
39 course[l].per=1.0*course[l].h/course[l].c;
40 }
41 qsort(course,n,sizeof(struct node),cmp);
42 ht=0;
43 ct=0;
44 maxs=0;
45 for (l=0;l<n;l++)
46 {
47 ht+=course[l].h;
48 ct+=course[l].c;
49 maxs=max(maxs,ht*ht-ht*ct-ct*ct);
50 }
51 printf("%lld\n",maxs);
52 }
53 return 0;
54 }
55 /*
56 5
57 1 4
58 2 5
59 6 3
60 7 2
61 4 4
62 */
最普通的方法，保证正确，用于测试其它程序是否正确：
1 #include <stdio.h>
2 #include <stdlib.h>
3
4 struct node
5 {
6 long h,c;
7 double per;
8 };
9 long n;
10 long long maxs=0;
11 struct node course[501];
12
13 int cmp(const void *a,const void *b)
14 {
15 return (*(struct node *)b).per-(*(struct node *)a).per;
16 }
17
18 long long max(long long a,long long b)
19 {
20 if (a>b)
21 return a;
22 else
23 return b;
24 }
25
26 void dfs(long d,long ht,long ct)
27 {
28 if (d==n)
29 return ;
30 if (ct!=0)
31 maxs=max(maxs,ht*ht-ht*ct-ct*ct);
32 dfs(d+1,ht+course[d].h,ct+course[d].c);
33 dfs(d+1,ht,ct);
34 }
35
36 int main()
37 {
38 long t,k,l;
39 scanf("%ld",&t);
40 for (k=1;k<=t;k++)
41 {
42 scanf("%ld",&n);
43 maxs=0;
44 for (l=0;l<n;l++)
45 {
46 scanf("%ld%ld",&course[l].h,&course[l].c);
47 course[l].per=1.0*course[l].h/course[l].c;
48 }
49 dfs(0,0,0);
50 printf("%lld\n",maxs);
51 }
52 return 0;
53 }
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
