sgu225 装压DP&位运算

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题意是：给出一个 n*n 的棋盘，问放置 k 的骑士相当于象棋中的马有多少中放置方法。。。

n等于10 是TLE ，于是 n=10 打表

用 dfs写的，运行时间长，但代码简单。。

开数组时用到滚动数组。。。

由于攻击范围是两行，所以是在三行之间做DP ，每种状态表示两行，这样上下三行之间的关系就两种状态直接的关系。。

相当于 f[ i ] [ j ] = opt(f[ j ][ t ]);

此题由于两行之间会攻击，所以两行状态合成一中状态时（就是用最长二十位的数来表示两行状态，10 位表示一行），合法的状态数就会减少。。可以预处理出所有合法状态。。。。

#include <cstdio>#include <cstring>#include <iostream>#define LL long long using namespace std;// 打表LL  mark[110]={1,100,4662,135040,2732909,41199404,481719518,4491423916,34075586550,213628255072,1120204619108,4961681221524,18715619717199,60541371615660,168976761361446,409191804533576,864172675710439,1599730843649564,2609262108838924,3770687313420780,4857550050070531,5616928666465104,5874943705896600,5604501518609804,4917655076255841,3999855946779732,3034690618677388,2156485957257040,1437827591264317,899278231344296,526753407546620,288274613750624,146990556682887,69626509814580,30542906352994,12366448408056,4604442057431,1569983914256,487876545370,137395261280,34831261750,7884855000,1578162590,275861904,41455966,5246412,543534,44244,2652,104,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};int state[30000]; //存取合法状态。。。。int fstate[(1<<20)+1000];int statenum =0 ;LL f[2][29000][52];int n,k;int bitnum(int i ){int sum =0;while(i!=0){  int  t = i&(-i);  sum++;  i-=t;}return sum ;}void init_state(){statenum =0;  int line1 ;  int line2 ;  for(int i=0 ;i<1<<(2*n);i++){      line2 = i>>n;      line1 = line2 << n;      line1 = i - line1;      if(line2&(line1<<2)){        continue;      }      if(line2&(line1>>2)){        continue;      }      statenum ++;      state[statenum]= i;      fstate[i] =statenum;  }}int maxidx []={0,1,5,6,9,14,18,26,32,42,51};void dfs(int r , int  i,int j, int  lev,int num){      int line1 =  j>>n;      int line2 =  line1<<n;          line2 = j -line2;      if(lev==n){        for(int t =num;t<=maxidx[n];t++){    int tmp = (line2<<n)+i;            f[r%2][fstate[tmp]][t] +=f[(r+1)%2][fstate[j]][t-num]; }return ;     }     int st = (line1>>1)|(line1<<1)|(line2>>2)|(line2<<2);          dfs(r,i,j,lev+1,num);     if(!(st&(1<<lev))){           dfs(r,i|(1<<lev),j,lev+1,num+1);     }}void  solve (){if(n==10){ cout << mark[k]<<endl; return ;}if(k>maxidx[n]){   cout << 0<<endl;;  return ;}        init_state();        memset(f,0,sizeof(f));           for(int  i=1;i<=statenum;i++){   if(i>(1<<n)){      break;   }   f[1][i][bitnum(state[i])]=1;}f[1][1][0]=1;for(int r=1;r<n;r++){    memset(f[(r+1)%2],0,sizeof(f[(r+1)%2]));        for(int j = 1;j<=statenum;j++){      dfs(r+1,0,state[j],0,0);    }}        LL ans =  0;for(int i=1;i<=statenum;i++){  ans +=  f[n%2][i][k];}cout <<ans<<endl;;}int main(){cin >> n>>k;solve();  return 0 ;}