题目链接

Just Jump

题意

有L+1个点，初始在第0个点上，要跳到第L个点，每次至少跳d格，也就是在点x至少要跳到x+d，且有m个限制
\((t_i, p_i)\)指跳第\(t_i\)次不能跳到\(p_i\)上

题解

设dp[i]表示从0跳到i且没有限制的方案数，可以预处理。对限制按t从小到大排序，\(g[i][0]\)表示通过了前i-1个限制，踩在了第i个限制上且1-i这些限制踩了偶数个，\(g[i][1]\)表示踩了奇数个限制的方案数，可以\(m^2\)递推，根据容斥原理加上踩偶次的方案数扣掉踩奇数次的方案数。

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
 
const int mx = 1e7+5;
const int mod = 998244353;  
ll dp[mx], fac[mx], invf[mx];
int g[3005][2];
int L, d, m;
 
int pow_mod(ll a, ll b) {
    ll ans = 1;
    while (b > 0) {
        if (b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b /= 2;
    }
    return ans;
}
 
struct Node {
    int t, p;
    bool operator < (Node other) const {
        return t < other.t;
    }
}node[3005];
 
int C(int x, int y) {
    if (x < y || y < 0 || x < 0) return 0;
    return (1LL * fac[x] * invf[y] % mod * invf[x-y]) % mod;
}
 
int calc(int i, int j) {
    if (1LL * (node[j].t-node[i].t)*d > node[j].p-node[i].p) return 0;
    return C(node[j].p-node[i].p-(node[j].t-node[i].t)*d + (node[j].t-node[i].t) - 1, node[j].t-node[i].t-1);
}
 
int main() {
    scanf("%d%d%d", &L, &d, &m);
    for (int i = 1; i <= m; i++) scanf("%d%d", &node[i].t, &node[i].p);
    sort(node+1, node+1+m);
    node[0].t = node[0].p = 0;
    dp[0] = 1;
    int sum = 0;
    for (int i = 1; i <= L; i++) {
        if (i-d >= 0) sum = (sum + dp[i-d]) % mod;
        dp[i] = sum;
    }
     
    fac[0] = invf[0] = 1;
    for (int i = 1; i <= L; i++) fac[i] = 1LL * fac[i-1] * i % mod;
    invf[L] = pow_mod(fac[L], mod-2);
    for (int i = L-1; i >= 1; i--) invf[i] = 1LL * invf[i+1] * (i+1) % mod;
 
    g[0][0] = 1; g[0][1] = 0;
    for (int i = 1; i <= m; i++) {
        for (int j = 0; j < i; j++) {
            g[i][0] = (g[i][0] + 1LL * g[j][1] * calc(j, i) % mod) % mod;
            g[i][1] = (g[i][1] + 1LL * g[j][0] * calc(j, i) % mod) % mod;
        }
    }
    ll ans = dp[L];
    for (int i = 1; i <= m; i++) {
        ans += 1LL * (g[i][0] - g[i][1]) * dp[L-node[i].p] % mod;
        ans = (ans % mod + mod) % mod;
    }
    printf("%lld\n", ans);
    return 0;
}