URAL 1707. Hypnotoad's Secret（树阵）

URAL 1707. Hypnotoad's Secret

题目链接

题意：这题设置的恶心不能多说。构造点和矩形。大概就是问每一个矩形里面是否包括点

思路：树状数组。把点排序，按y轴，在按x轴。在按询问，这样每次遇到一个点就在对应的扫描线上加。遇到查询就询问出左边到这个点位置的，就能预处理出每一个点左下角包括的点的个数，然后每一个矩形再利用容斥原理去搞一下就可以

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;

typedef long long ll;

const int N = 1050005;
const int M = 5005;

const ll MOD = 200904040930 + 33;

struct Point {
int x, y;
bool isop;
Point() {}
Point(int x, int y, bool isop) {
this->x = x;
this->y = y;
this->isop = isop;
}
} p
;

bool cmp(Point a, Point b) {
if (a.y == b.y) {
if (a.x == b.x)
return a.isop < b.isop;
return a.x < b.x;
}
return a.y < b.y;
}

int pn;

int n, m;

struct OP {
int a0, b0, c0, d0;
int da, db, dc, dd;
int q;
void read() {
scanf("%d%d%d%d%d%d%d%d%d", &a0, &b0, &c0, &d0, &da, &db, &dc, &dd, &q);
}
} op[350];

#define lowbit(x) (x&(-x))

int bit[M];

void add(int x, int v) {
while (x <= n) {
bit[x] += v;
x += lowbit(x);
}
}

int get(int x) {
int ans = 0;
while (x) {
ans += bit[x];
x -= lowbit(x);
}
return ans;
}

int get(int l, int r) {
return get(r) - get(l - 1);
}

typedef pair<int, int> pii;
#define MP(a,b) make_pair(a,b)

map<pii, int> cnt;
int save[350];
ll mi7[350];

int main() {
mi7[0] = 1;
for (int i = 1; i < 350; i++)
mi7[i] = mi7[i - 1] * 7 % MOD;
while (~scanf("%d%d", &n, &m)) {
cnt.clear();
pn = 0;
memset(bit, 0, sizeof(bit));
int s0, t0, ds, dt, k;
while (m--) {
scanf("%d%d%d%d%d", &s0, &t0, &ds, &dt, &k);
for (int i = 0; i < k; i++) {
p[pn++] = Point(s0, t0, false);
s0 = ((s0 + ds) % n + n) % n;
t0 = ((t0 + dt) % n + n) % n;
}
}
scanf("%d", &m);
for (int i = 0; i < m; i++) {
op[i].read();
int a0 = op[i].a0, b0 = op[i].b0, c0 = op[i].c0, d0 = op[i].d0;
int da = op[i].da, db = op[i].db, dc = op[i].dc, dd = op[i].dd;
int q = op[i].q;
for (int j = 0; j < q; j++) {
int a = min(a0, b0), b = max(a0, b0), c = min(c0, d0), d = max(c0, d0);
p[pn++] = Point(a - 1, c - 1, true);
p[pn++] = Point(b, c - 1, true);
p[pn++] = Point(a - 1, d, true);
p[pn++] = Point(b, d, true);
a0 = ((a0 + da) % n + n) % n;
b0 = ((b0 + db) % n + n) % n;
c0 = ((c0 + dc) % n + n) % n;
d0 = ((d0 + dd) % n + n) % n;
}
}
sort(p, p + pn, cmp);
for (int i = 0; i < pn; i++) {
if (p[i].isop) cnt[MP(p[i].x, p[i].y)] = get(1, p[i].x + 1);
else add(p[i].x + 1, 1);
}
for (int i = 0; i < m; i++) {
int a0 = op[i].a0, b0 = op[i].b0, c0 = op[i].c0, d0 = op[i].d0;
int da = op[i].da, db = op[i].db, dc = op[i].dc, dd = op[i].dd;
int q = op[i].q;
for (int j = 0; j < q; j++) {
int a = min(a0, b0), b = max(a0, b0), c = min(c0, d0), d = max(c0, d0);
int tmp = cnt[MP(b, d)] - cnt[MP(a - 1, d)] - cnt[MP(b, c - 1)] + cnt[MP(a - 1, c - 1)];
if (tmp == 0) save[j] = 0;
else save[j] = 1;
a0 = ((a0 + da) % n + n) % n;
b0 = ((b0 + db) % n + n) % n;
c0 = ((c0 + dc) % n + n) % n;
d0 = ((d0 + dd) % n + n) % n;
}
if (q <= 20) {
for (int j = 0; j < q; j++)
printf("%d", save[j]);
printf("\n");
} else {
ll out = 0;
for (int j = 0; j < q; j++)
out = (out + mi7[j] * save[j]) % MOD;
printf("%lld\n", out);
}
}
}
return 0;
}