Codeforces 722E [DP]

Solution

以前做过一道题是这样的

• 给定n∗m的网格图和k个点，求不定过某些点从(1,1)走到(n,m)的方案数。

这个的话直接O(k2)DP就好了。
这道题也差不多，因为电池的衰减只有O(log2s)种，所以DP时记录一下当前经过的点的个数就好了。

#include <cstdio>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;const int N = 101010;const int M = 2020;const int MOD = 1000000007;typedef long long ll;inline char get(void) {    static char buf[100000], *S = buf, *T = buf;    if (S == T) {        T = (S = buf) + fread(buf, 1, 100000, stdin);        if (S == T) return EOF;    }    return *S++;}inline void read(int &x) {    static char c; x = 0;    for (c = get(); c < '0' || c > '9'; c = get());    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';}int n, m, k, s, lim, ans;struct Point {    int x, y;    Point (int _x = 0, int _y = 0):x(_x), y(_y) {}    inline friend bool operator <(const Point &a, const Point &b) {        return a.x == b.x ? a.y < b.y : a.x < b.x;    }};Point a[M];int fac[N << 1], inv[N << 1];int dp[N][100];inline int Pow(int a, int b) {    int c = 1;    while (b) {        if (b & 1) c = (ll)c * a % MOD;        b >>= 1; a = (ll)a * a % MOD;    }    return c;}inline int Inv(int x) {    return Pow(x, MOD - 2);}inline int C(int n, int m) {    return (ll)fac[n] * inv[m] % MOD * inv[n - m] % MOD;}inline int Calc(int n, int m) {    return C(n + m, n);}int main(void) {    read(n); read(m); read(k); read(s);    inv[1] = 1; lim = 25;    for (int i = 2; i <= 200000; i++)        inv[i] = (ll)(MOD - MOD / i) * inv[MOD % i] % MOD;    fac[0] = inv[0] = 1;    for (int i = 1; i <= 200000; i++) {        fac[i] = (ll)fac[i - 1] * i % MOD;        inv[i] = (ll)inv[i - 1] * inv[i] % MOD;    }    for (int i = 1; i <= k; i++) {        read(a[i].x); read(a[i].y);    }    sort(a + 1, a + k + 1);    if (a[k].x == n && a[k].y == m) s -= s / 2;    else a[++k] = Point(n, m);    if (a[1].x != 1 || a[1].y != 1) {        s *= 2; a[++k] = Point(1, 1);    }    sort(a + 1, a + k + 1);    dp[1][0] = 1;    for (int i = 2; i <= k; i++) {        dp[i][1] = Calc(a[i].x - 1, a[i].y - 1);        for (int d = 2; d <= lim; d++) {            for (int j = 1; j < i; j++) {                if (a[j].y > a[i].y) continue;                dp[i][d] += (ll)dp[j][d - 1] * Calc(a[i].x - a[j].x, a[i].y - a[j].y) % MOD;                dp[i][d] -= (ll)dp[j][d] * Calc(a[i].x - a[j].x, a[i].y - a[j].y) % MOD;                dp[i][d] = (dp[i][d] % MOD + MOD) % MOD;            }        }    }    ans = 0;    for (int i = 1; i <= lim; i++) {        s -= s / 2;        ans = ans + (ll)(dp[k][i] - dp[k][i + 1] + MOD) * s % MOD;        ans %= MOD;    }    cout << (ll)ans * Inv(Calc(n - 1, m - 1)) % MOD << endl;    return 0;}