题目:http://www.lydsy.com/JudgeOnline/problem.php?id=1151
状压DP,枚举前面4个,使得环型变线型。
#include<cstdio> #include<cstdlib> #include<iostream> #include<fstream> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<queue> #include<stack> #include<map> #include<utility> #include<set> #include<bitset> #include<vector> #include<functional> #include<deque> #include<cctype> #include<climits> #include<complex> //#include<bits/stdc++.h>适用于CF,UOJ,但不适用于poj using namespace std; typedef long long LL; typedef double DB; typedef pair<int,int> PII; typedef complex<DB> CP; #define mmst(a,v) memset(a,v,sizeof(a)) #define mmcy(a,b) memcpy(a,b,sizeof(a)) #define re(i,a,b) for(i=a;i<=b;i++) #define red(i,a,b) for(i=a;i>=b;i--) #define fi first #define se second #define m_p(a,b) make_pair(a,b) #define SF scanf #define PF printf #define two(k) (1<<(k)) template<class T>inline T sqr(T x){return x*x;} template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;} template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;} const DB EPS=1e-9; inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;} const DB Pi=acos(-1.0); inline int gint() { int res=0;bool neg=0;char z; for(z=getchar();z!=EOF && z!=‘-‘ && !isdigit(z);z=getchar()); if(z==EOF)return 0; if(z==‘-‘){neg=1;z=getchar();} for(;z!=EOF && isdigit(z);res=res*10+z-‘0‘,z=getchar()); return (neg)?-res:res; } inline LL gll() { LL res=0;bool neg=0;char z; for(z=getchar();z!=EOF && z!=‘-‘ && !isdigit(z);z=getchar()); if(z==EOF)return 0; if(z==‘-‘){neg=1;z=getchar();} for(;z!=EOF && isdigit(z);res=res*10+z-‘0‘,z=getchar()); return (neg)?-res:res; } const int maxN=10000; const int maxC=50000; #define wei(v,k) (((v)>>(k-1))&1) int N,C; struct Tkid { int pos,afraid,love; inline void input() { int i,F,L; pos=gint()+4;F=gint();L=gint(); afraid=0;re(i,1,F){int t=gint();if(pos>N && t<=4)t+=N;afraid|=two(pos-t);} love=0;re(i,1,L){int t=gint();if(pos>N && t<=4)t+=N;love|=two(pos-t);} } inline int happy(int state){return ((~state)&love) || (state & afraid);} }kid[maxC+100]; int F[maxN+100][two(5)+10]; int ans; int main() { freopen("bzoj1151.in","r",stdin); freopen("bzoj1151.out","w",stdout); int i,j,l; N=gint();C=gint(); re(i,1,C)kid[i].input(); int S; re(S,0,two(4)-1) { mmst(F,-1); F[4][S]=0; int head=1,tail; re(i,5,N)re(j,0,two(5)-1)re(l,0,1) { while(head<=C && kid[head].pos<i)head++; if(F[i-1][j]==-1)continue; int k,t; k=(j*2+l)%two(5); t=F[i-1][j]; for(tail=head;tail<=C && kid[tail].pos==i;tail++)if(kid[tail].happy(k))t++; upmax(F[i][k],t); } re(i,N+1,N+4)re(j,0,two(5)-1) { while(head<=C && kid[head].pos<i)head++; if(F[i-1][j]==-1)continue; int k,t; k=(j*2+wei(S,5+N-i))%two(5); t=F[i-1][j]; for(tail=head;tail<=C && kid[tail].pos==i;tail++)if(kid[tail].happy(k))t++; upmax(F[i][k],t); } re(j,0,two(5)-1)upmax(ans,F[N+4][j]); } cout<<ans<<endl; return 0; }
时间: 2024-10-12 12:32:09