【Luogu】P3052摩天大楼里的奶牛(遗传算法乱搞)
一道状压题,但今天闲来无事又用遗传乱搞了一下。
设了一个DNA数组,DNA[i]记录第i个物品放在哪个组里。适应度是n-这个生物的组数+1.
交配选用的是轮盘赌和单亲繁殖——0.3的几率单点变异。(事实上有性生殖我似乎写不出来……代码量略大)
种群大小开到了400,在vijos上繁殖了2050代,下数据自己测也是对的。
然而只有84分
这究竟是为什么啊 下数据自己测是没错的啊………………
疯了
代码和数据先放到这里,以后再改吧
#include<cstdio>
#include<cctype>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<ctime> inline double random(double from,double to){
return rand()*1.0/RAND_MAX*(to-from)+from;
} inline long long read(){
long long num=,f=;
char ch=getchar();
while(!isdigit(ch)){
if(ch=='-') f=-;
ch=getchar();
}
while(isdigit(ch)){
num=num*+ch-'';
ch=getchar();
}
return num*f;
} int n;
int w[],W;
int ans=0x7fffffff; struct Creative{
int DNA[],s[];
int cnt,score;
Creative(){ cnt=;score=; memset(s,,sizeof(s)); }
Creative Create(){
Creative New;
for(register int i=;i<=n;++i){
if(New.cnt==||random(,)<=0.5){
New.cnt++;
New.DNA[i]=New.cnt;
New.s[New.cnt]+=w[i];
}
else{
int j,tot=;
Label:
tot++;
j=random(,New.cnt);
if(New.s[j]+w[i]>W&&tot<=n*) goto Label;
if(!New.s[j]&&tot<=n+) goto Label;
if(tot>=n*){
New.cnt++;
New.DNA[i]=New.cnt;
New.s[New.cnt]+=w[i];
continue;
}
New.DNA[i]=j;
New.s[j]+=w[i];
}
}
if(New.cnt) ans=std::min(ans,New.cnt);
New.score=n-New.cnt+;
return New;
}
}; struct Biome{
Creative e[];
int tot;
void Clear(bool flag){
tot=;
for(int i=;i<;++i){
if(flag) e[i]=e[i].Create();
if(e[i].cnt) tot+=e[i].score;
if(e[i].cnt==) i--;
}
}
int Find(){
int sum=,limit=random(,tot);
for(register int i=;i<;++i){
sum+=e[i].score;
if(sum>=limit) return i;
}
return ;
}
Creative Multi(int x){
Creative New=e[x];
if(random(,)>0.3) return New;
int i=random(,n+),pos=New.DNA[i];
New.s[pos]-=w[i];
if(!New.s[pos]){
New.cnt--;
New.score++;
}
int j,tot=;
Label:
tot++;
j=random(,n);
if(New.s[j]+w[i]>W&&tot<=n*) goto Label;
if(!New.s[j]&&tot<=n+) goto Label;
if(tot>=n*){
New.cnt++;
New.DNA[i]=New.cnt;
New.score--;
New.s[New.cnt]+=w[i];
}
else{
New.DNA[i]=j;
if(!New.s[j]){
New.cnt++;
New.score--;
}
New.s[j]+=w[i];
ans=std::min(ans,New.cnt);
}
return New;
}
}Old,New; inline bool cmp(Creative a,Creative b){ return a.score>b.score; } int main(){
srand(time(NULL));
n=read(),W=read();
for(int i=;i<=n;++i) w[i]=read();
Old.Clear();
for(register int T=;T<=;++T){
Old.Clear();
for(register int i=;i<;++i) New.e[i]=Old.Multi(Old.Find());
std::sort(Old.e+,Old.e+,cmp);
std::sort(New.e+,New.e+,cmp);
for(register int i=;i<=;++i) Old.e[+i]=New.e[i];
}
/*for(int i=1;i<=100;++i,printf("\n")){
for(int j=1;j<=n;++j) printf("%d ",Old.e[i].DNA[j]);
printf(">>>%d %d",Old.e[i].cnt,Old.e[i].score);
}*/
printf("%d",ans);
return ;
}
18 100000000
37597832
24520955
100000000
18509980
29141223
20287969
14028193
33097076
12116817
53439913
10216168
32891936
43952038
13463011
4056577
4646046
10153053
37881213
下一个准备用遗传搞的题是猫狗大战。
【Luogu】P3052摩天大楼里的奶牛(遗传算法乱搞)的更多相关文章
- 【Luogu】P3052摩天大楼里的奶牛(状压DP)
参见ZHT467的题解. f[i]表示在i这个集合下的最少分组数和当前组最少的容量. 从1到(1<<n)-1枚举i,对于每个i枚举它的子奶牛,然后重载运算符计算. 代码如下 #includ ...
- 洛谷P3052 [USACO12MAR]摩天大楼里的奶牛 [迭代加深搜索]
题目传送门 摩天大楼里的奶牛 题目描述 A little known fact about Bessie and friends is that they love stair climbing ra ...
- 洛谷P3052 [USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper
P3052 [USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper 题目描述 A little known fact about Bessie and friends is ...
- [USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper
洛谷题目链接:[USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper 题目描述 A little known fact about Bessie and friends is ...
- 【题解】Luogu P3052 【USACO12】摩天大楼里的奶牛Cows in a Skyscraper
迭代加深搜索基础 题目描述 A little known fact about Bessie and friends is that they love stair climbing races. A ...
- LUOGU P3052 [USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper
题目描述 A little known fact about Bessie and friends is that they love stair climbing races. A better k ...
- P3052 [USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper
题目描述 给出n个物品,体积为w[i],现把其分成若干组,要求每组总体积<=W,问最小分组.(n<=18) 输入格式: Line 1: N and W separated by a spa ...
- 洛谷 P3052 [USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper
题目描述 A little known fact about Bessie and friends is that they love stair climbing races. A better k ...
- P3052 [USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper 状压dp
这个状压dp其实很明显,n < 18写在前面了当然是状压.状态其实也很好想,但是有点问题,就是如何判断空间是否够大. 再单开一个g数组,存剩余空间就行了. 题干: 题目描述 A little k ...
随机推荐
- IT之家学院:使用CMD命令行满速下载百度云
转自:https://www.toutiao.com/a6545305189685920259/?tt_from=android_share&utm_campaign=client_share ...
- GoAccess参数选项
GoAccess - 1.2 Usage: goaccess [filename] [ options ... ] [-c][-M][-H][-q][-d][...]The following opt ...
- MySQL存储过程(更新指定字段的数据)
mysql存储过程示例: USE 数据库名称;DROP PROCEDURE IF EXISTS 数据库名称.存储过程名称;delimiter $$CREATE PROCEDURE 数据库名称.存储过程 ...
- React学习实例总结,包含yeoman安装、webpack构建
1.安装yeoman 在安装nodeJs的基础上,输入命令:npm install -g yo grunt-cli bower,安装yeoman,grunt,bowerify 安装完成后,输入命令:y ...
- 状态压缩---区间dp第一题
标签: ACM 题目 Gappu has a very busy weekend ahead of him. Because, next weekend is Halloween, and he is ...
- 讲课笔记1——meta标签、表格标签
图片属性:src(source): 图片的来源(路径),可以放置本地图片,也可以放网上的图片的url地址 [相对路径: ./:当前目录 ../:跳出当前目录,到上一 ...
- 稳定性 耗时 gc 过长问题排查 和工具
自己的另外一篇: http://www.cnblogs.com/fei33423/p/7805186.html 偶有耗时抖动? gc 也有长耗时? fullgc 也是? 有同学反馈 swap 可能导致 ...
- UIControlEvent
UIControlEventTouchDown = 1 << 0, // 手指落在按钮的一瞬间触发UIControlEventTouchDownRepeat ...
- STATIC 和 CLASS
STATIC 和 CLASS 由 王巍 (@ONEVCAT) 发布于 2015/01/28 Swift 中表示 “类型范围作用域” 这一概念有两个不同的关键字,它们分别是 static 和 class ...
- 使用python制作查询火车票工具
使用python脚本实现查询火车票信息的效果图如下: 实现的代码: # coding: utf-8 """命令行火车票查看器 Usage: tickets [-gdtkz ...