【UOJ #204】【APIO 2016】Boat

http://uoj.ac/problem/204
確定要離散化的，先離散化出\(O(n)\)個取值區間。
設\(f(i,j)\)表示第\(i\)所學校派出的划艇數量在\(j\)區間中。
\(f(i,j)=\sum\limits_{k=0}^{i-1}\left(\sum\limits_{t=1}^{j-1}f(k,t)\right)\times Cal(k+1,i,j)\)
\(Cal(l,r,j)\)表示\([l,r)\)中的每所學校要否則不派出划艇，要否則派出數量在\(j\)區間中的划艇，第\(r\)所學校必定要派出數量在\(j\)區間中的划艇，且知足划艇數遞增的方案個數。
假設\([l,r)\)中只有\(m\)所學校能知足派出數量在\(j\)區間中的划艇，設\(j\)區間的大小爲\(l\)，那麼\(Cal(l,r,j)=\sum\limits_{i=1}^{m+1}{l\choose i}\times{m\choose i-1}={l+m\choose m+1}\)。
利用這個組合數，再記錄一下dp的前綴和，時間複雜度\(O(n^3)\)。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;

const int N = 503;
const int p = 1000000007;

int l[N], a[N], b[N], H[N << 1], cnt = 0, n, f[N][N << 1], ni[N];

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d%d", a + i, b + i);
        H[++cnt] = a[i];
        H[++cnt] = ++b[i];
    }
    
    stable_sort(H + 1, H + cnt + 1);
    cnt = unique(H + 1, H + cnt + 1) - H;
    
    for (int i = 1; i <= n; ++i) {
        a[i] = lower_bound(H + 1, H + cnt, a[i]) - H;
        b[i] = lower_bound(H + 1, H + cnt, b[i]) - H;
    }
    
    cnt -= 2;
    for (int i = 1; i <= cnt; ++i)
        l[i] = H[i + 1] - H[i];
    
    ni[1] = 1;
    for (int i = 2; i <= n; ++i)
        ni[i] = 1ll * (p - p / i) * ni[p % i] % p;
    
    for (int i = 0; i <= cnt; ++i) f[0][i] = 1;
    for (int i = 1; i <= n; ++i) {
        for (int j = a[i], up = b[i]; j < up; ++j) {
            int C = l[j], r = l[j], c = 1;
            for (int k = i - 1; k >= 0; --k) {
                (f[i][j] += 1ll * f[k][j - 1] * C % p) %= p;
                if (a[k] <= j && j < b[k]) {
                    ++r; ++c;
                    C = 1ll * C * r % p * ni[c] % p;
                }
            }
        }
        
        for (int j = 2; j <= cnt; ++j)
            (f[i][j] += f[i][j - 1]) %= p;
    }
    
    int ans = 0;
    for (int i = 1; i <= n; ++i) (ans += f[i][cnt]) %= p;
    printf("%d\n", ans);
    return 0;
}