【UOJ #204】【APIO 2016】Boat

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)=sumlimits_{k=0}^{i-1}left(sumlimits_{t=1}^{j-1}f(k,t) ight) imes 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)=sumlimits_{i=1}^{m+1}{lchoose i} imes{mchoose i-1}={l+mchoose 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
", ans);
	return 0;
}