procedure2012
It's not worth it to know you're not worth it!

[关键字]:随机化算法 模拟退火

[题目大意]:http://221.192.240.123:8586/JudgeOnline/showproblem?problem_id=1677

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[分析]:本来还想把它独立做出来,结果碰着这么一道“RP完全问题”,在冥思苦想了几个小时的搜索、贪心、动态规划都无果后,终于放弃了。看完题解发现居然是一道模拟退火的题。先用搜索找出第三类点所有可能的分类,然后对每一类先随便找一个A、B机器的执行序列,然后随机A中的两个点交换顺序,在随机B中的两个点交换顺序,看是否更优如果更优就保存当前顺序并继续执行否则退回上一顺序,大概执行2000多次后,就有很大几率得到最优解(看RP当然也有概率分析)。

[代码]:

View Code
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<ctime>
#include<cmath>
using namespace std;
#define MAXN 100
#define INF 1000000

int N,top1,topa,topb,ans=INF;
int c[MAXN],a[MAXN],b[MAXN];
int s1[MAXN],sa[MAXN],sb[MAXN],ssa[MAXN],ssb[MAXN];

bool cmp1(int A,int B)
{
if (b[A]>b[B]||(b[A]==b[B]&&a[A]<=a[B]))return true;
else return false;
}

bool cmp2(int A,int B)
{
if (a[A]>a[B]||(a[A]==a[B]&&b[A]<=b[B]))return true;
else return false;
}

void Solve()
{
topa=topb=0;
int ta=0,tb=0;
for (int i=1;i<=N;i++)
if (c[i]==1) sa[++topa]=i,ta+=a[i]; else sb[++topb]=i,tb+=b[i];
sort(sa+1,sa+topa+1,cmp1);
sort(sb+1,sb+topb+1,cmp2);
int t1=ta,t2=tb;
int Max=INF,v1,v2;
for (int i=1;i<=2000;i++)
{
ta=t1,memcpy(ssa,sa,sizeof(sa));
tb=t2,memcpy(ssb,sb,sizeof(sb));
if (topa) v1=rand()%topa+1,v2=rand()%topa+1,swap(sa[v1],sa[v2]);
if (topb) v1=rand()%topb+1,v2=rand()%topb+1,swap(sb[v1],sb[v2]);
int temp=0;
for (int j=1;j<=topa;j++)
{
temp+=a[sa[j]];
tb=max(temp,tb)+b[sa[j]];
}
temp=0;
for (int j=1;j<=topb;j++)
{
temp+=b[sb[j]];
ta=max(temp,ta)+a[sb[j]];
}
temp=max(ta,tb);
if (temp>Max) memcpy(sa,ssa,sizeof(sa)),memcpy(sb,ssb,sizeof(sb));
ans=min(temp,ans);
Max=min(Max,temp);
}
}

void Dfs(int k)
{
if (k>top1)
{
Solve();
return ;
}
int limit=20000;
c[s1[k]]=1;
if (rand()%32767<limit) Dfs(k+1);
c[s1[k]]=2;
if (rand()%32767<limit) Dfs(k+1);
}

int main()
{
scanf("%d",&N);
for (int i=1;i<=N;i++)
{
scanf("%d%d%d",&c[i],&a[i],&b[i]);
if (c[i]==3) s1[++top1]=i;
}
srand((unsigned)time(NULL));
Dfs(1);
printf("%d\n",ans);
return 0;
}



posted on 2012-02-20 18:44  procedure2012  阅读(1017)  评论(3编辑  收藏  举报