[luogu P6144] [USACO20FEB]Help Yourself P

我丢： https://www.luogu.com.cn/problem/P6144

先考虑 k = 1
考虑DP
首先将所有的线段按照左端点排序，考虑 $f[r]$表示最右以$r$结尾的线段集合的连通块的数量的和.

插入一个线段$[l,r]$，

$对于右端点在[1,l-1]的每种情况，都可以使它们的连通块加1,再加到f[r]里面$
$对于右端点在[l,r]的每种情况，可以直接加到f[r]里面$
$对于右端点在[r+1,n]的f[i],要全部\ast 2(选当前线段或不选)$

也就是要一个数据结构，支持区间*2,区间求和，显然线段树

对于k != 1, 可以考虑把它的 [0，k]次的结果分别维护，计算的时候用二项式定理合并就好了

$就是对于[1,l-1]的要把x^k->(x+1{)}^k再加到f[r]里，这里可以用二项式定理把[0,k]次方的结果乘上二项式系数再加起来就好了$

就是线段树每个点要维护各个次方的结果

看代码好理解一点

code:

#include
#define N 400005
#define int long long
#define mod 1000000007
using namespace std;
struct haha {
	int a[13];
};
struct A {
	int l, r;
} a[N];
int cmp(A x, A y) {
	return x.l < y.l;
}
int ha[N << 3][12], tag[N << 3], n, k, C[25][25];
void update(int rt) {
	for(int i = 0; i <= k; i ++)
	 	ha[rt][i] = (ha[rt << 1][i] + ha[rt << 1 | 1][i]) % mod;
}
void pushdown(int rt) {
	if(tag[rt] == 1) return;
	tag[rt << 1] *= tag[rt], tag[rt << 1 | 1] *= tag[rt], tag[rt << 1 | 1] %= mod, tag[rt << 1] %= mod;
	int t = 1;
	for(int i = 0; i <= k; i ++) 
		ha[rt << 1][i] *= tag[rt], ha[rt << 1 | 1][i] *= tag[rt], ha[rt << 1][i] %= mod, ha[rt << 1 | 1][i] %= mod;
	tag[rt] = 1;
}	
void add(int rt, int l, int r, int x, haha b) { 
	if(l == r) {
		for(int i = 0; i <= k; i ++) ha[rt][i] = (ha[rt][i] + b.a[i]) % mod;
		return;
	}
	pushdown(rt);
	int mid = (l + r) >> 1;
	if(x <= mid) add(rt << 1, l, mid, x, b);
	else add(rt << 1 | 1, mid + 1, r, x, b);
	update(rt);
}
void mul(int rt, int l, int r, int L, int R) {
	if(L > R) return;
	if(L <= l && r <= R) {
		tag[rt] = tag[rt] * 2 % mod;
		int t = 1;
		for(int i = 0; i <= k; i ++)
			ha[rt][i] = ha[rt][i] * 2 % mod;
		return;
	}
	pushdown(rt);
	int mid = (l + r) >> 1;
	if(L <= mid) mul(rt << 1, l, mid, L, R);
	if(R > mid) mul(rt << 1 | 1, mid + 1, r, L, R);
	update(rt);
}
haha query(int rt, int l, int r, int L, int R) { 
	if(L > R) {
		haha b;
		for(int i = 0; i <= k; i ++) b.a[i] = 0;
		return b;
	}
	if(L <= l && r <= R) {
		haha d;
		for(int i = 0; i <= k; i ++) d.a[i] = ha[rt][i];
		return d;
	}
	pushdown(rt);
	int mid = (l + r) >> 1; haha ret;
	ret.a[0] = 1;
	for(int i =  0; i <= k; i ++) ret.a[i] = 0;
	if(L <= mid) {
		haha b = query(rt << 1, l, mid, L, R);
		for(int i = 0; i <= k; i ++) ret.a[i] += b.a[i], ret.a[i] %= mod;
	} 
	if(R > mid) {
		haha b = query(rt << 1 | 1, mid + 1, r, L, R);
		for(int i = 0; i <= k; i ++) ret.a[i] += b.a[i], ret.a[i] %= mod;
	}
	return ret;
}
signed main() {
	scanf("%lld%lld", &n, &k); int lim = 2 * n;
	for(int i = 0; i <= lim * 4; i ++) ha[i][0] = 0	;
	for(int i = 0; i <= k; i ++) C[i][0] = 1;
	for(int i = 1; i <= k; i ++)
		for(int j = 1; j <= i; j ++)
			C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
	for(int i = 1; i <= n; i ++) scanf("%lld%lld", &a[i].l, &a[i].r);
	sort(a + 1, a + 1 + n, cmp);
	haha d;
	for(int i = 0; i <= k; i ++) d.a[i] = 0;
	d.a[0] = 1;
	add(1, 0, lim, 0, d);
	for(int tt = 1; tt <= n; tt ++) {
		int l = a[tt].l, r = a[tt].r;
		haha b = query(1, 0, lim, 0, l - 1);
		haha c;
		for(int i = 0; i <= k; i ++) c.a[i] = 0;
		for(int i = 0; i <= k; i ++){
			int ret = 0;
			for(int j = 0; j <= i; j ++)
				c.a[i] += b.a[j] * C[i][j] % mod, c.a[i] %= mod;
		}
		b = query(1, 0, lim, l, r - 1);
		for(int i = 0; i <= k; i ++) c.a[i] = (c.a[i] + b.a[i]) % mod;
		add(1, 0, lim, r, c);
		mul(1, 0, lim, r + 1, lim);
	}
	printf("%lld", query(1, 0, lim, 0, lim).a[k]);
	return 0;
}

这一场就这题值得写一下QWQ