poj1185 [NOI2001炮兵阵地]

题目链接

状压DP

本来如果考虑所有情况应该开hh[n][2^10][2^10]表示i行在i-1的状态为j，i-2的状态为k的最大个数

但是由于每行中的人互相限制所以在m=10时只有60种情况

空间就可以满足，时间也可以满足了

 #include<algorithm>
 #include<iostream>
 #include<cstdlib>
 #include<cstring>
 #include<cstdio>
 #include<string>
 #include<cmath>
 #include<ctime>
 #include<queue>
 #include<stack>
 #include<map>
 #include<set>
 #define rre(i,r,l) for(int i=(r);i>=(l);i--)
 #define re(i,l,r) for(int i=(l);i<=(r);i++)
 #define Clear(a,b) memset(a,b,sizeof(a))
 #define inout(x) printf("%d",(x))
 #define douin(x) scanf("%lf",&x)
 #define strin(x) scanf("%s",(x))
 #define LLin(x) scanf("%lld",&x)
 #define op operator
 #define CSC main
 typedef unsigned long long ULL;
 typedef const int cint;
 typedef long long LL;
 using namespace std;
 void inin(int &ret)
 {
     ret=;int f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=;ch=getchar();}
     while(ch>=''&&ch<='')ret*=,ret+=ch-'',ch=getchar();
     ret=f?-ret:ret;
 }
 int s[],x,shu[];
 int n,m,a[];
 int hh[][][];
 char ss[][];
 void dfs(int xx,int ss,int sh)
 {
     if(xx>m)
     {
         s[++x]=ss;shu[x]=sh;
         return ;
     }
     dfs(xx+,ss,sh);
     dfs(xx+,ss|(<<xx),sh+);
 }
 int CSC()
 {
     freopen("in.in","r",stdin);
     freopen("out.out","w",stdout);
     inin(n),inin(m);
     dfs(,,);
     re(i,,n)
     {
         strin(ss[i]+);
         re(j,,m)if(ss[i][j]=='H')a[i]|=(<<j);
     }
     re(i,,x)if(!(a[]&s[i]))
         hh[][i][]=shu[i];
     re(i,,x)
         if(!(a[]&s[i]))
         re(j,,x)
             if(!(s[i]&s[j]))
             hh[][i][j]=max(hh[][i][j],hh[][j][]+shu[i]);
     re(r,,n)
         re(i,,x)if(!(s[i]&a[r]))
         re(j,,x)if(!(s[j]&a[r-]))if(!(s[i]&s[j]))
             re(k,,x)if(!(s[k]&a[r-]))if(!(s[j]&s[k]))if(!(s[i]&s[k]))
                 hh[r][i][j]=max(hh[r][i][j],hh[r-][j][k]+shu[i]);
     int ans=;
     re(i,,x)re(j,,x)
         ans=max(ans,hh[n][i][j]);
     printf("%d",ans);
     return ;
 }