Luogu P1377 M国王

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表示裸状压DP
f[i][j][k]j表示第i行的国王位置，k表示i行和i行前共有k个国王

f[i][j][k+num[j]]=∑f[i−1][o][k]}
ans=∑2n−1i=0f[n][i][m]

Then,AC

#1AC0ms/13839kB#2AC0ms/13839kB#3AC0ms/13839kB#4AC0ms/13839kB#5AC1ms/13839kB

#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>#include<math.h>#include<vector>#include<map>#include<queue>#include<time.h>#include<fstream>#include<string>#include<set>#include<list>#include<stdlib.h>#define fr(i,a,b) for(int i=a,_end_=b;i<=_end_;i++)#define fd(i,a,b) for(int i=a,_end_=b;i>=_end_;i--)#define frei(s) freopen(s,"r",stdin)#define freo(s) freopen(s,"w",stdout)#define ll long long#define uns unsignedusing namespace std;#define rt return#define gc getchar()#define ln putchar('\n')ll lowbit(ll x){    rt x&(-x);}int read(){    int s=0,k=1;    char c=getchar();    while((c<'0'||c>'9')&&c!='-')c=gc;    if(c=='-'){        k=-1;        c=gc;    }    while(c>=48&&c<='9'){        s=s*10+c-48;        c=gc;    }    rt s*k;}int exgcd(int a,int b,int &x,int &y)  {      if(b==0)      {          x=1;        y=0;          return a;      }      int d=exgcd(b,a%b,x,y);      int temp=x;      x=y;      y=temp-a/b*y;      return d;  }  double log(double x,double y){    rt log10(x)/log10(y);}ll gcd(ll x,ll y){    rt y?gcd(y,x%y):x;}ll P(ll x,ll y)//x>=y{    ll r=1;    fr(i,1,y)        r*=(x-i+1);    rt r;}ll P(ll x,ll y,ll modnum)//x>=y{    ll r=1;    fr(i,1,y)        r=r*(x-i+1)%modnum;    rt r;}int n,m,ans,num[1<<9],f[10][1<<9][100];int main(){#ifndef ONLINE_JUDGE    freopen("","r",stdin);    freopen("","w",stdout);#endif    n=read();    m=read();    fr(i,1,(1<<n)-1)        num[i]=1+num[i-(i&(-i))];//计算有几个国王    f[0][0][0]=1;//边界    fr(i,1,n)        fr(j,0,(1<<n)-1)            if(!((j&(j<<1))|(j&(j>>1))))//判断可不可以这样放                fr(o,0,(1<<n)-1)                    if(!((j&(o<<1))|(j&(o>>1))|(j&o)))//判断是否冲突                        fr(k,0,m)                            f[i][j][k+num[j]]+=f[i-1][o][k];    fr(i,0,(1<<n)-1)        ans+=f[n][i][m];//累加每一个情况    printf("%d\n",ans);    rt 0;}

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