BZOJ 1801 中国象棋(DP)

如果考虑每一行的状态的话，是会有后效性的，解决办法就是改变状态的表示。

注意到每一行只能放0个，1个，或2个炮。那么可以依据这个转移来定义状态。

定义dp[i][j][k]表示前i行有j列0个炮，k列1个炮，m-j-k列两个炮的方法数。

则转移方程是显然的。我们剩下的只需要记忆化搜索一下即可。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-9
# define MOD 9999973
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=105;
//Code begin...

int dp[N][N][N], n, m;

int dfs(int x, int a, int b){
    if (~dp[x][a][b]) return dp[x][a][b];
    if (a+b>m||a<0||b<0) return 0;
    if (x==1) {
        if (a+b!=m||b>2) return 0;
        if (b==1) return dp[x][a][b]=m;
        if (b==2) return dp[x][a][b]=m*(m-1)/2;
        return dp[x][a][b]=1;
    }
    int res=(dfs(x-1,a,b)+(a+1)*dfs(x-1,a+1,b-1)%MOD+(b+1)*dfs(x-1,a,b+1)%MOD)%MOD;
    res=(res+(LL)(a+1)*b*dfs(x-1,a+1,b)%MOD+(LL)(a+2)*(a+1)/2*dfs(x-1,a+2,b-2)%MOD+(LL)(b+2)*(b+1)/2*dfs(x-1,a,b+2)%MOD)%MOD;
    return dp[x][a][b]=res;
}
int main ()
{
    int ans=0;
    mem(dp,-1);
    scanf("%d%d",&n,&m);
    FOR(i,0,m) FOR(j,0,m-i) ans=(ans+dfs(n,i,j))%MOD;
    printf("%d
",ans);
    return 0;
}