hdu 4832 Chess(计数+dp)

题目链接：hdu 4832 Chess

题目大意：略。（注意King只能走周围8格）

解题思路：将水平和竖直分开考虑，l[i]表示竖直上走i步不出界的种数，r[i]表示水平上走i步不出界的种数，然后枚举水平竖直走的步数（相加为k），并且要乘以组合数。因为确定步数了但是还要考虑先后的关系。
处理步数的时候，开一个二维数组dp[i][j],表示i步，位置在j的种数，j为偏移，j-k为负数表示在起始点左/上的|j−k|位置，j-k为正数表示在起点右/下的|j−k|的位置上。

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;
typedef long long ll;

const int N = 1010;
const ll MOD = 9999991;

int n, m, k, x, y;
ll l[N], r[N], g[N][N*2], c[N][N];

void cat (ll* a, int x, int t) {
    memset(g, 0, sizeof(g));
    g[0][k] = 1;

    int up = k-(x-1);
    int down = k+(t-x);

    for (int i = 1; i <= k; i++) {
        for (int j = up; j <= down; j++) {
            if (g[i-1][j]) {
                if (j != up) {
                    g[i][j-1] = (g[i][j-1] + g[i-1][j]) % MOD;
                    if (j != up+1)
                        g[i][j-2] = (g[i][j-2] + g[i-1][j]) % MOD;
                }

                if (j != down) {
                    g[i][j+1] = (g[i][j+1] + g[i-1][j]) % MOD;
                    if (j != down-1)
                        g[i][j+2] = (g[i][j+2] + g[i-1][j]) % MOD;
                }
            }
            a[i-1] = (a[i-1] + g[i-1][j]) % MOD;
        }
    }

    for (int i = up; i <= down; i++)
        a[k] = (a[k] + g[k][i]) % MOD;
}

void input () {
    for (int i = 0; i < N; i++) {
        c[i][0] = c[i][i] = 1;
        for (int j = 0; j < i; j++)
            c[i][j] = (c[i-1][j-1] + c[i-1][j]) % MOD;
    }
}

void init () {
    //scanf("%d%d%d%d%d", &n, &m, &k, &x, &y);
    cin >> n >> m >> k >> x >> y;
    memset(l, 0, sizeof(l));
    memset(r, 0, sizeof(r));
    cat(l, x, n);
    cat(r, y, m);
}

ll solve () {
    ll ans = 0;
    for (int i = 0; i <= k; i++)
        ans = (ans + (l[i] * r[k-i]) % MOD * c[k][i])%MOD;

    return ans;
}

int main () {
    input();
    int cas;
    cin >> cas;
    for (int i = 1; i <= cas; i++) {
        init();
        //printf("Case #%d:\n%lld\n", i, solve());
        cout << "Case #" << i << ":" << endl;
        cout << solve() << endl;
    }
    return 0;
}