海盗分金问题SQL求解（贪心算法）

问题

经济学上有个“海盗分金”模型：是说5个海盗抢得100枚金币，他们按抽签的顺序依次提方案：首先由1号提出分配方案，然后5人表决，超过半数同意方案才被通过，否则他将被扔入大海喂鲨鱼，依此类推,假设海盗是足够聪明的先利己再伤人，最后方案是怎样的？

网上百度来的的代码

with a as 
(select 101 - rownum  n from dual connect by rownum <102),
max_one as 
(select max(n) max1 from a),
max_two as 
(select /*+leading(p2,p1) use_nl(p1) */ p2.n max2,p1.n max1 
 from a p1,a p2
 where p1.n+p2.n=100
 and p1.n=(select max1 from max_one)
 and rownum=1),
max_three as 
(select /*+leading(p3,p2,p1) use_nl(p2) use_nl(p1)*/ p3.n max3,p2.n max2,p1.n max1
 from a p1,a p2,a p3,max_two
 where p1.n+p2.n+p3.n=100
 and sign(p2.n-max2)+sign(p1.n-max1)>=0
 and rownum=1),
max_four as 
(select /*+leading(p4,p3,p2,p1) use_nl(p3) use_nl(p2) use_nl(p1)*/ p4.n max4,p3.n max3,p2.n max2,p1.n max1
 from a p1,a p2,a p3,a p4,max_three
 where p1.n+p2.n+p3.n+p4.n=100
 and sign(p3.n-max3)+sign(p2.n-max2)+sign(p1.n-max1)>0
 and rownum=1),
five as 
(select /*+leading(p5,p4,p3,p2,p1) use_nl(p4) use_nl(p3) use_nl(p2) use_nl(p1)*/ p5.n n5, p4.n n4,p3.n n3,p2.n n2,p1.n n1
 from a p1,a p2,a p3,a p4,a p5,max_four
 where p1.n+p2.n+p3.n+p4.n+p5.n=100
 and sign(p4.n-max4)+sign(p3.n-max3)+sign(p2.n-max2)+sign(p1.n-max1)>=0
 and rownum=1)
select * from five;

严格筛选数据优化后

with a as 
(select 101 - rownum  n from dual connect by rownum <102),
max_one as 
(select max(n) max1 from a),
max_two as 
(select /*+leading(max_one,p2,p1) use_nl(p2) use_nl(p1) */ p2.n max2,p1.n max1 
 from a p1,a p2,max_one
 where p1.n+p2.n=100
 and p1.n>=max1
 and rownum=1),
max_three as 
(select /*+leading(max_two,p3,p2,p1) use_nl(max_two) use_nl(p2) use_nl(p1)*/ p3.n max3,p2.n max2,p1.n max1
 from a p1,a p2,a p3,max_two
 where p1.n+p2.n+p3.n=100
 and p3.n+p2.n<=100
 and case when p2.n > max2 then 1 else -1 end +
     case when p1.n > max1 then 1 else -1 end >= 0
 and rownum=1),
max_four as 
(select /*+leading(max_three,p4,p3,p2,p1) use_nl(max_three) use_nl(p3) use_nl(p2) use_nl(p1)*/ p4.n max4,p3.n max3,p2.n max2,p1.n max1
 from a p1,a p2,a p3,a p4,max_three
 where p1.n+p2.n+p3.n+p4.n=100
 and p4.n+p3.n <= 100
 and p4.n+p3.n+p2.n <= 100
 and case when p3.n > max3 then 1 else -1 end +
     case when p2.n > max2 then 1 else -1 end +
     case when p1.n > max1 then 1 else -1 end >= 0
 and rownum=1),
five as 
(select /*+leading(max_four,p5,p4,p3,p2,p1) use_nl(p5) use_nl(p4) use_nl(p3) use_nl(p2) use_nl(p1)*/ p5.n n5, p4.n n4,p3.n n3,p2.n n2,p1.n n1
 from a p1,a p2,a p3,a p4,a p5,max_four
 where p1.n+p2.n+p3.n+p4.n+p5.n=100
 and p5.n+p4.n <= 100
 and p5.n+p4.n+p3.n <= 100
 and p5.n+p4.n+p3.n+p2.n <= 100
 and case when p4.n > max4 then 1 else -1 end + 
     case when p3.n > max3 then 1 else -1 end +
     case when p2.n > max2 then 1 else -1 end +
     case when p1.n > max1 then 1 else -1 end >= 0
 and rownum=1)
select * from five;

结果