• CF R274 Div2 E Riding in a Lift DP


    先预处理出能到当前点的区间,然后通过前缀和求得当前值即可。

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #include <climits>
    #include <iostream>
    #include <string>
    
    using namespace std;
     
    #define MP make_pair
    #define PB push_back
    #define lson rt << 1, l, mid
    #define rson rt << 1 | 1, mid + 1, r
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef vector<int> VI;
    typedef pair<int, int> PII;
    typedef pair<double, double> PDD;
    const int INF = INT_MAX / 3;
    const double eps = 1e-8;
    const double DINF = 1e60;
    const int maxn = 5005;
    const LL mod = 1e9 + 7;
    int n, a, b, k;
    int f[maxn][maxn], sum[maxn][maxn];
    int lbd[maxn], rbd[maxn];
    
    int mabs(int x) {
        return x < 0 ? -x : x;
    }
    
    int main() {
        cin >> n >> a >> b >> k;
        for(int i = 1; i <= n; i++) {
            lbd[i] = INF; rbd[i] = -INF;
        }
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                if(mabs(j - i) < mabs(j - b)) {
                    lbd[i] = min(lbd[i], j);
                    rbd[i] = max(rbd[i], j);
                }
            }
        }
        lbd[b] = rbd[b] = b;
        f[0][a] = 1;
        for(int i = 1; i <= n; i++) sum[0][i] = sum[0][i - 1] + f[0][i];
        for(int i = 1; i <= k; i++) {
            for(int j = 1; j <= n; j++) {
                int l = lbd[j], r = rbd[j];
                f[i][j] = (sum[i - 1][r] - sum[i - 1][l - 1] + mod) % mod;
                if(lbd[j] <= j && rbd[j] >= j) 
                    f[i][j] = (f[i][j] - f[i - 1][j] + mod) % mod;
            }
            for(int j = 1; j <= n; j++) 
                sum[i][j] = (sum[i][j - 1] + f[i][j]) % mod;
        }
        cout << sum[k][n] << endl;
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/rolight/p/4041631.html
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