二维数点问题 (从线段树到CDQ分治)

ps当然还有的二维树状数组 这里暂时不提

star

http://acm.hdu.edu.cn/showproblem.php?pid=1541
统计 x y 到 0 0 有多少星星
排序 按x y 升序 排 前面只影响后面 离散化 树状数组 统计
HDU 星星 这道题 算是简单题 数据范围也没有看 自己在胡诌离散化 处理大矩阵数据了
找板子题 没有找到 以下代码 带离散化 可以处理的矩阵相当大了

#include
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;

int n, cnt;
struct node {
	int x, y;
	bool operator < (const node& a) const {
		if(x != a.x) return x < a.x;
		else return y < a.y;
	}
} a[maxn];
int c[maxn], b[maxn], ans[maxn];
void add(int x, int y) {
	for(; x <= n; x += (x & (-x))) c[x] += y;
}

int sum(int x) {
	int res = 0;
	for(; x; x -= (x & (-x))) res += c[x];
	return res;
}

int main() {
	while(cin >> n) {
		memset(c, 0, sizeof c);
		memset(ans, 0, sizeof ans);	
		for(int i = 1; i <= n; i ++)
			cin >> a[i].x >> a[i].y, b[i] = a[i].y;
		sort(a + 1, a + 1 + n);
		sort(b + 1, b + 1 + n);
		int cnt = unique(b + 1, b + 1 + n) - b - 1;
		int now;
		for(int i = 1; i <= n; i ++) {
			int pos = lower_bound(b + 1, b + 1 + cnt, a[i].y) - b;
			now = sum(pos + 1);
			ans[now] ++;
			add(pos + 1, 1);
		}
		for(int i = 0; i < n; i ++) {
			printf("%d\n", ans[i]);
		}
	}
	return 0;
}

P2163 [SHOI2007]园丁的烦恼

本来想着二维数组 然后不对劲啊
CDQ分治嚎啊
如果你理解的CDQ 分治 这题并不难

#include
using namespace std;
typedef long long ll;
const int maxn = 3e6 + 10;
const int INF = 0x3f3f3f3f;
typedef pair<int, int> P;

int n, m, cnt;
struct node {
	int x, y, op, id;
	bool operator < (const node &b) const {
		return x == b.x ? (y == b.y ? op < b.op : y < b.y) : x < b.x;
	}
} que[maxn], tmp[maxn];
int ans[maxn];
void cdq(int L, int R) {
	if(L == R) return ;
	int mid = L + R >> 1;
	cdq(L, mid), cdq(mid + 1, R);
	int l = L, r = mid + 1, pos = L, tot = 0;
	while(l <= mid && r <= R) {
		if(que[l].y <= que[r].y)
			tot += !que[l].op, tmp[pos ++] = que[l ++];
		else
			ans[que[r].id] += tot, tmp[pos ++] = que[r ++];
	}
	while(l <= mid) tmp[pos ++] = que[l ++];
	while(r <= R) ans[que[r].id] += tot, tmp[pos ++] = que[r ++];
	for(int i = L; i <= R; i ++) que[i] = tmp[i];
}

int main() {
	scanf("%d %d", &n, &m);
	for(int i = 1, x, y; i <= n; i ++) {
		scanf("%d %d", &x, &y);
		que[++ cnt] = node {x, y, 0, 0};
	}
	int tot = 0;
	for(int i = 1, a, b, c, d; i <= m; i ++) {
		scanf("%d %d %d %d", &a, &b, &c, &d);
		que[++ cnt] = node {c, d, 1, ++ tot};
		que[++ cnt] = node {c, b - 1, 1, ++ tot};
		que[++ cnt] = node {a - 1, d, 1, ++ tot};
		que[++ cnt] = node {a - 1, b - 1, 1, ++ tot};
	}
	sort(que + 1, que + 1 + cnt);
	cdq(1, cnt);
	for(int i = 1; i + 3 <= tot; i += 4)
		printf("%d\n", ans[i] - ans[i + 1] - ans[i + 2] + ans[i + 3]);
	return 0;
}

CDQ 陌上花开 动态逆序对

这里是三维偏序
https://blog.csdn.net/qq_40831340/article/details/99587234

P4390 [BOI2007]Mokia 摩基亚

依然是三维偏序 注意坐标 要加一 题目给的是 在格子中数量

#include
using namespace std;
typedef long long ll;
const int maxn = 4e6 + 10;
const int INF = 0x3f3f3f3f;
typedef pair<int, int> P;

int n, m, cnt;
struct node { // 1是询问 ; 0 是需改
	int x, y, z, op, id, val;
} que[maxn], tmp[maxn];
int ans[maxn], c[maxn];

bool cmp(const node &a, const node &b) {
	if(a.x != b.x) return a.x < b.x;
	if(a.y != b.y) return a.y < b.y;
	if(a.z != b.z) return a.z < b.z;
	return a.op < b.op;
}

void add(int x, int y) {
	for(; x <= n; x += (x & (-x))) c[x] += y;
}

void add(int x) {
	for(; x <= n; x += (x & (-x))) c[x] = 0;
}

int sum(int x) {
	int res = 0;
	for(; x; x -= (x & (-x))) res += c[x];
	return res;
}

void cdq(int L, int R) {
	if(L == R) return ;
	int mid = L + R >> 1;
	cdq(L, mid), cdq(mid + 1, R);
	int l = L, r = mid + 1, pos = L, tot = 0;
	while(l <= mid && r <= R) {
		if(que[l].y <= que[r].y) {
			if(!que[l].op) add(que[l].z, que[l].val);
			tmp[pos ++] = que[l ++];
		} else {
			if(que[r].op) ans[que[r].id] += sum(que[r].z);
			tmp[pos ++] = que[r ++];
		}
	}
	while(r <= R) {
		if(que[r].op) ans[que[r].id] += sum(que[r].z);
		tmp[pos ++] = que[r ++];
	}
	while(l <= mid) {
		if(!que[l].op) add(que[l].z, que[l].val);
		tmp[pos ++] = que[l ++];
	}

	for(int i = L; i <= R; i ++) {
		if(i <= mid && !que[i].op) add(que[i].z);
		que[i] = tmp[i];
	}
}

int main() {
	scanf("%d %d", &m, &n);
	++ n;
	int tot = 0, tim = 0;
	int x1, y1, x2, y2, val, op;
	while(scanf("%d", &op) && op != 3) {
		if(op == 1) {
			tim ++;
			scanf("%d %d %d", &x1, &y1, &val);
			que[++ cnt] = node {x1 + 1, y1 + 1, tim, 0, 0, val};
		} else {
			scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
			x2 += 1, y2 += 1;
			que[++ cnt] = node {x2, y2, tim, 1, ++tot, 0};
			que[++ cnt] = node {x2, y1, tim, 1, ++tot, 0};
			que[++ cnt] = node {x1, y2, tim, 1, ++tot, 0};
			que[++ cnt] = node {x1, y1, tim, 1, ++tot, 0};
		}
	}
	n = tim;
	sort(que + 1, que + 1 + cnt, cmp);
	cdq(1, cnt);
	for(int i = 1; i + 3 <= tot; i += 4)
		printf("%d\n", ans[i] - ans[i + 1] - ans[i + 2] + ans[i + 3]);
	return 0;
}