本篇博客将会从贡献的思想认识前缀和.

首先,我们回顾一下一维前缀和和二维前缀和:

//一维
for (int i = 1; i <= n; ++i)
    s[i] = s[i - 1] + a[i];
//二维
for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
        s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i];

当然,也可以这么写:

//一维
for (int i = 1; i <= n; ++i)
    s[i] += s[i - 1];
//二维
for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
        f[i][j] += f[i - 1][j];
for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
        f[i][j] += f[i][j - 1];

二维前缀和的第一种写法实际上就是容斥原理,不在我们本篇讨论的范围内.

我们主要讲第二种:

我们讲每个f[i][j]初始化为一个值a[i][j],如果我们要求二维前缀和,我们考虑a[i][j]对哪些前缀和产生了贡献.

 答案就是上图中的蓝色部分,我们要使得a(i, j)对蓝色部分造成贡献,可以分成两部分,对列造成贡献.

 然后对于每个列,我们往下推

 问题就转化成了,如何让一个点,对这个点后面的直线产生贡献,答案就是一维的前缀和,直接向想要产生贡献的方向累加即可.

同理,我们想要求左下角方向的"前缀和",我们也可以先向左边累加,然后向下边累加

 同样的,对于更高维的前缀和,我们采取相同的做法.

CF372B Counting Rectangles is Fun

题目传送门

 我们先枚举(i, j, i1, j1),如果这是个全0矩阵,我们Q(i, j, i1, j1)++,然后我们假设f(i, j, i1, j1)为这个矩形内部的全0矩阵的个数,考虑每个子矩阵(i, j, i1, j1)产生的贡献.就是红色部分,我们直接按照这些方向累加即可:

#include 
#define int long long
#define IOS ios::sync_with_stdio(false), cin.tie(0) 
#define ll long long 
// #define double long double
#define ull unsigned long long 
#define PII pair 
#define PDI pair 
#define PDD pair 
#define debug(a) cout << #a << " = " << a << endl 
#define point(n) cout << fixed << setprecision(n)
#define all(x) (x).begin(), (x).end() 
#define mem(x, y) memset((x), (y), sizeof(x)) 
#define lbt(x) (x & (-x)) 
#define SZ(x) ((x).size()) 
#define inf 0x3f3f3f3f 
#define INF 0x3f3f3f3f3f3f3f3f
namespace nqio{const unsigned R = 4e5, W = 4e5; char *a, *b, i[R], o[W], *c = o, *d = o + W, h[40], *p = h, y; bool s; struct q{void r(char &x){x = a == b && (b = (a = i) + fread(i, 1, R, stdin), a == b) ? -1 : *a++;} void f(){fwrite(o, 1, c - o, stdout); c = o;} ~q(){f();}void w(char x){*c = x;if (++c == d) f();} q &operator >>(char &x){do r(x);while (x <= 32); return *this;} q &operator >>(char *x){do r(*x); while (*x <= 32); while (*x > 32) r(*++x); *x = 0; return *this;} template<typename t> q&operator>>(t &x){for (r(y),s = 0; !isdigit(y); r(y)) s |= y == 45;if (s) for (x = 0; isdigit(y); r(y)) x = x * 10 - (y ^ 48); else for (x = 0; isdigit(y); r(y)) x = x * 10 + (y ^ 48); return *this;} q &operator <<(char x){w(x);return *this;}q &operator<< (char *x){while (*x) w(*x++); return *this;}q &operator <<(const char *x){while (*x) w(*x++); return *this;}template<typename t> q &operator<< (t x) {if (!x) w(48); else if (x < 0) for (w(45); x; x /= 10) *p++ = 48 | -(x % 10); else for (; x; x /= 10) *p++ = 48 | x % 10; while (p != h) w(*--p);return *this;}}qio; }using nqio::qio;
using namespace std;
const int N = 1e6 + 10, M = (1 << 21) + 5, MOD = 1e9 + 7;
int n, m, f[M], qp[M] = {1};
void work(int x) {
	for (int j = 0; j < 21; ++j)
		for (int i = 0; i < (1 << 21); ++i)
			if (i >> j & 1)
				f[i] = ((f[i] + f[i ^ (1ll << j)] * x % MOD) % MOD + MOD) % MOD;
}
void solve() {
	qio >> n >> m;
	for (int i = 1; i <= n; ++i) qp[i] = (qp[i - 1] << 1ll) % MOD;
	for (int i = 1; i <= n; ++i) {
		int k, st = 0;
		qio >> k;
		for (int j = 1, x; j <= k; ++j) qio >> x, --x, st |= (1ll << x);
		++f[st];
	}
	work(1);
	for (int i = 0; i < (1 << 21); ++i) f[i] = (qp[f[i]] - 1) % MOD;
	work(-1);
	qio << f[(1 << m) - 1] << '\n';
}
signed main()  {
	// IOS;
	int T = 1;
	// qio >> T;
	while (T--) solve();
}

除此之外,前缀和还可以用来解决偏序问题,不过不在此篇讨论范围内.