BZOJ_P1087&Codevs_P2451 [SCOI2005]互不侵犯King(状态压缩DP)

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Time Limit: 10 Sec Memory Limit: 162 MB

Submit: 2385 Solved: 1399

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Description

在N×N的棋盘里面放K个国王，使他们互不攻击，共有多少种摆放方案。国王能攻击到它上下左右，以及左上左下右上右下八个方向上附近的各一个格子，共8个格子。

Input

只有一行，包含两个数N，K （ 1 <=N <=9, 0 <= K <= N * N）

Output

方案数。

Sample Input

3 2

Sample Output

16

HINT

Source

#include<cstdio>
#include<iostream>
using namespace std;
#define N 10
int n,k,cnt;long long f
[N*N][1<<N];long long ans;
int m[1<<N];
inline int calc(long long x){
int res=0;
for(int i=0;i<10;i++,x>>=1) if(x&1) res++;
return res;
}
inline bool can1(int S){//单行是否可行
for(int i=0;i<N;i++)
if(S&(1<<i)&&S&(1<<(i+1))) return false;
return true;
}
inline bool can2(int S1,int S2){//两行是否可行
for(int i=0;i<N;i++)
if((1<<i)&S1)
if((1<<(i+1)&S2)||(1<<(i-1)&S2)||((1<<i)&S2)) return false;
return true;
}
int main(){
scanf("%d%d",&n,&k);cnt=(1<<n);
for(int i=0;i<cnt;i++) m[i]=calc(i);//计算每位1的数量
for(int S=0;S<cnt;S++){
if(m[S]<=k) f[0][m[S]][S]=1;
}
for(int i=1;i<n;i++){
for(int j=0;j<=k;j++)
for(int S1=0;S1<cnt;S1++){//上一状态
if(m[S1]>j||(!can1(S1))) continue;
for(int S2=0;S2<cnt;S2++){//当前状态
if(can1(S2))
if(j+m[S2]<=k&&can2(S1,S2)){
f[i][m[S2]+j][S2]+=f[i-1][j][S1];
}
}
}
}
for(int S=0;S<cnt;S++){
ans+=f[n-1][k][S];
}
printf("%lld",ans);
return 0;
}

哦~~这题数据辣么小让我打上40s的表

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
#define N 10
int n,k,cnt;long long f
[N*N][1<<N];long long ans;
int m[1<<N];
inline int calc(long long x){
int res=0;
for(int i=0;i<10;i++,x>>=1) if(x&1) res++;
return res;
}
inline bool can1(int S){
for(int i=0;i<N;i++)
if(S&(1<<i)&&S&(1<<(i+1))) return false;
return true;
}
inline bool can2(int S1,int S2){
for(int i=0;i<N;i++)
if((1<<i)&S1)
if((1<<(i+1)&S2)||(1<<(i-1)&S2)||((1<<i)&S2)) return false;
return true;
}
int main(){
freopen("std.out","w",stdout);
for(int p=1;p<=9;p++){
for(int q=0;q<p*p;q++){
n=p,k=q;memset(f,0,sizeof(f));
cnt=(1<<n);ans=0;
for(int i=0;i<cnt;i++) m[i]=calc(i);
for(int S=0;S<cnt;S++){
if(m[S]<=k) f[0][m[S]][S]=1;
}
for(int i=1;i<n;i++){
for(int j=0;j<=k;j++)
for(int S1=0;S1<cnt;S1++){//上一状态
if(m[S1]>j||(!can1(S1))) continue;
for(int S2=0;S2<cnt;S2++){//当前状态
if(can1(S2))
if(j+m[S2]<=k&&can2(S1,S2)){
f[i][m[S2]+j][S2]+=f[i-1][j][S1];
}
}
}
}
for(int S=0;S<cnt;S++){
ans+=f[n-1][k][S];
}
printf("%I64d,",ans);
}
for(int i=p*p+1;i<81;i++) printf("0,");
}
return 0;
}

然后可以提交介个 ps:内存有点大啊喂，缩小点就进BZOJ第一页了

#include<cstdio>
long long f[]={1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,9,16,8,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,16,78,140,79,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,25,228,964,1987,1974,978,242,27,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,36,520,3920,16834,42368,62266,51504,21792,3600,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,49,1020,11860,85275,397014,1220298,2484382,3324193,2882737,1601292,569818,129657,18389,1520,64,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,64,1806,29708,317471,2326320,12033330,44601420,119138166,229095676,314949564,305560392,204883338,91802548,25952226,4142000,281571,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,81,2968,65240,962089,10087628,77784658,450193818,1979541332,6655170642,17143061738,33787564116,50734210126,57647295377,49138545860,31122500764,14518795348,4959383037,1237072414,224463798,29275410,2673322,163088,6150,125,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
int main(){
int n,k;
scanf("%d%d",&n,&k);
printf("%lld",f[(n-1)*81+k-(n-1)]);
return 0;
}