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
作者:九命猫幺
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博客出处:http://www.cnblogs.com/yongestcat/
欢迎转载,转载请标明出处。
如果你觉得本文还不错,对你的学习带来了些许帮助,请帮忙点击右下角的推荐