http://uoj.ac/problem/204

肯定要离散化的，先离散化出\(O(n)\)个取值区间。

设\(f(i,j)\)表示第\(i\)所学校派出的划艇数量在\(j\)区间中。

\(f(i,j)=\sum\limits_{k=0}^{i-1}\left(\sum\limits_{t=1}^{j-1}f(k,t)\right)\times Cal(k+1,i,j)\)

\(Cal(l,r,j)\)表示\([l,r)\)中的每所学校要不然不派出划艇，要不然派出数量在\(j\)区间中的划艇，第\(r\)所学校一定要派出数量在\(j\)区间中的划艇，且满足划艇数递增的方案个数。

假设\([l,r)\)中只有\(m\)所学校能满足派出数量在\(j\)区间中的划艇，设\(j\)区间的大小为\(l\)，那么\(Cal(l,r,j)=\sum\limits_{i=1}^{m+1}{l\choose i}\times{m\choose i-1}={l+m\choose m+1}\)。

利用这个组合数，再记录一下dp的前缀和，时间复杂度\(O(n^3)\)。

#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;

}