[组合数 DP] HDU 4532 湫秋系列故事——安排座位

一种典型的组合数DP 

f[i][j]表示前i种数值排列有j个不合法间隙的方案数

然后枚举第i个分成几块k 有l块塞入j个不合法空隙中进行转移

答案即为f
[0]

不过还要乘以排列数

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#define cl(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long ll;

inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=55;
const ll P=1e9+7;

int n;
int cnt
,sum
;
ll dp
[555];
ll C[555][555],fac[555];

inline void Pre(){
C[0][0]=1;
for (int i=1;i<=500;i++){
C[i][0]=1;
for (int j=1;j<=i;j++)
(C[i][j]=C[i-1][j-1]+C[i-1][j])%=P;
}
fac[0]=1;
for (int i=1;i<=500;i++)
fac[i]=fac[i-1]*i%P;
}

int main(){
int T,test=0;
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
Pre();
read(T);
while (T--){
read(n);
for (int i=1;i<=n;i++) read(cnt[i]),sum[i]=sum[i-1]+cnt[i];
dp[1][cnt[1]-1]=1;
for (int i=2;i<=n;i++)
for (int j=0;j<=sum[i-1]+1;j++)
if (dp[i-1][j])
for (int k=1;k<=cnt[i];k++)
for (int l=0;l<=k;l++)
(dp[i][j-l+cnt[i]-k]+=dp[i-1][j]*C[j][l]%P*C[sum[i-1]+1-j][k-l]%P*C[cnt[i]-1][k-1]%P)%=P;
ll ret=dp
[0];
for (int i=1;i<=n;i++)
ret*=fac[cnt[i]],ret%=P;
printf("Case %d: %lld\n",++test,ret);
cl(dp);
}
return 0;
}