luogu 2051 中国象棋

Posted 2021-01-10 asdic

非常好的dp，锻炼思维

f[i][j][k] 前i行有j列放1，k列放2

#include<bits/stdc++.h>
#define int long long
#define rep(i,x,y) for(register int i=x;i<=y;i++)
#define dec(i,x,y) for(register int i=x;i>=y;i--)
using namespace std;
const int mod=999983;
const int N=150;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch==‘-‘)f=-1;ch=getchar();}
    while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    return x*f;}
int f[N][N][N],n,m;
inline int C(int x){return x*(x-1)/2;}
signed main(){
    freopen("count.in","r",stdin);
    freopen("count.out","w",stdout);
    n=read();m=read();
    f[0][0][0]=1;
    for(int i=0;i<n;i++)
    for(int j=0;j<=m;j++)
    for(int k=0;j+k<=m;k++){
        //不放 
        f[i+1][j][k]=(f[i+1][j][k]+f[i][j][k])%mod;
        //放1在0
        if(m-j-k>=1) f[i+1][j+1][k]=(f[i+1][j+1][k]+f[i][j][k]*(m-j-k))%mod;
        //放1在1
        if(j>=1) f[i+1][j-1][k+1]=(f[i+1][j-1][k+1]+f[i][j][k]*j)%mod; 
        //放2在0在0 
        if(m-j-k>=2) f[i+1][j+2][k]=(f[i+1][j+2][k]+f[i][j][k]*C(m-j-k))%mod;
        //放2在0在1
        if(m-j-k>=1&&j>=1) f[i+1][j][k+1]=(f[i+1][j][k+1]+f[i][j][k]*(m-j-k)*j)%mod;
        //放2在1在1
        if(j>=2) f[i+1][j-2][k+2]=(f[i+1][j-2][k+2]+f[i][j][k]*C(j))%mod;}
    int ans=0;
    for(int i=0;i<=m;i++)
    for(int j=0;i+j<=m;j++)
        ans=(ans+f[n][i][j])%mod;
    printf("%lld
",ans);return 0;
}