P1896 [SCOI2005]互不侵犯

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和上一题差不多，注意初始化与判断即可

#include<bits/stdc++.h>
using namespace std;
long long int n,kk,tot,can[4096],dp[11][1096][121],cnt[4096];
long long ans;
int main()
{
    cin>>n>>kk;
    for(int i=0;i<=(1<<n)-1;i++)
    if((!(i&(i<<1)))){can[++tot]=i;    
    int q=i;
    while(q){if(q%2==1){cnt[i]++;}q>>=1;}
    }
    for(int j=1;j<=tot;j++)dp[1][can[j]][cnt[can[j]]]=1;
    for(int i=2;i<=n;i++)
    for(int j=1;j<=tot;j++)
    for(int k=1;k<=tot;k++)
    {if((can[j]&can[k])||((can[j]<<1)&can[k])||((can[j]>>1)&can[k]))continue;
    for(int t=0;t<=kk;t++)dp[i][can[j]][t+cnt[can[j]]]+=dp[i-1][can[k]][t];}
    for(int i=1;i<=tot;i++)ans+=dp[n][can[i]][kk];
    cout<<ans;
}

标签：4096,互不侵犯,int,P1896,long,kk,tot,ans,SCOI2005
来源： https://www.cnblogs.com/SFWR-YOU/p/11261565.html