BZOJ 4806 炮

BZOJ 4806 炮

以前做过类似的
显然每一行每一列都不可以多于2个炮
DP状态$f[i][j][k]$表示第$i$行有j行有1个炮,有k行有2个炮
边界是$f[0][0][0] = 1$
然后分6种情况大力讨论一下用组合数解决即可.
CODE

#include <iostream>
#include <cstdio>
#define rep(i,x,p) for(register int i = x;i <= p;++ i)
#define sep(i,x,p) for(register int i = x;i >= p;-- i)
#define gc getchar()
#define pc putchar
const int maxN = 100 + 7;
const int mod = 999983;

long long f[2][maxN][maxN];
int n,m;

inline int read() {
    int x = 0,f = 1;char c = gc;
    while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
    return x * f;
}

inline int C(int n) {
    return n * ( n - 1 ) / 2;
}

int main() {
    f[0][0][0] = 1;
    int n,m;
    n = read();m = read();
    rep(i,1,n) {
        int x = i % 2,q = x ? 0 : 1;
        rep(j , 0 , m) {
            for(register int k = 0;k + j <= m;++ k) {
                f[x][j][k] = f[q][j][k];
                if(j >= 1) f[x][j][k] += f[q][j - 1][k] * (m - (j - 1) - k);
                if(k >= 1) f[x][j][k] += f[q][j + 1][k - 1] * (j + 1);
                if(k >= 2) f[x][j][k] += f[q][j + 2][k - 2] * C(j + 2);
                if(j >= 2) f[x][j][k] += f[q][j - 2][k] * C(m - j - k + 2);
                if(k >= 1) f[x][j][k] += f[q][j][k - 1] * j * (m - j - k + 1);
                f[x][j][k] %= mod;
            }
        }
    }
    long long ans = 0,x = n % 2;
    rep(j , 0 , m) {
        for(register int k = 0;k + j <= m;++ k) 
            ans += f[x][j][k];
        ans %= mod;
    }
    printf("%lld\n", ans);
    return 0;
}