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bzoj1151

时间:2015-07-15 22:31:48      阅读:189      评论:0      收藏:0      [点我收藏+]

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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。
技术分享
#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;
    }
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bzoj1151

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原文地址:http://www.cnblogs.com/maijing/p/4649503.html

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