This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "hexagon.h"
#include <stdlib.h>
#include <vector>
using namespace std;
typedef vector<int> vi;
const int N = 200000, MD = 1000000007, V2 = 500000004, V6 = 166666668;
int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
unsigned int X = 12345;
int rand_() {
return (X *= 3) >> 1;
}
int dx[] = { 1, 1, 0, -1, -1, 0 };
int dy[] = { 0, 1, 1, 0, -1, -1 };
int choose2(int n) {
return (long long) n * (n - 1) % MD * V2 % MD;
}
int choose3(int n) {
return (long long) n * (n - 1) % MD * (n - 2) % MD * V6 % MD;
}
long long cross(int x1, int y1, int x2, int y2) {
return (long long) x1 * y2 - (long long) x2 * y1;
}
int hh[N], ll[N], ii[N], xx[N][2], yy[N], n, n_;
void add(int x1, int y1, int x2, int y2) {
if (y1 < y2)
xx[n_][0] = x1, xx[n_][1] = x2, yy[n_] = y1, n_++;
else if (y1 > y2)
xx[n_][0] = x2, xx[n_][1] = x1, yy[n_] = y2, n_++;
}
void sort(int *ii, int l, int r) {
while (l < r) {
int i = l, j = l, k = r, i_ = ii[l + rand_() % (r - l)], tmp;
while (j < k) {
int c = yy[ii[j]] != yy[i_] ? yy[ii[j]] - yy[i_] : (xx[ii[j]][0] + xx[ii[j]][1]) - (xx[i_][0] + xx[i_][1]);
if (c == 0)
j++;
else if (c < 0) {
tmp = ii[i], ii[i] = ii[j], ii[j] = tmp;
i++, j++;
} else {
k--;
tmp = ii[j], ii[j] = ii[k], ii[k] = tmp;
}
}
sort(ii, l, i);
l = k;
}
}
int *ej[N], eo[N];
void append(int i, int j) {
int o = eo[i]++;
if (o >= 2 && (o & o - 1) == 0)
ej[i] = (int *) realloc(ej[i], o * 2 * sizeof *ej[i]);
ej[i][o] = j;
}
int dfs(int p, int i, int d) {
int i_ = i >> 1, s = i & 1, o, x;
x = (long long) d * (xx[ii[i_ << 1 | 1]][s] - xx[ii[i_ << 1 | 0]][s] + 1) % MD;
for (o = eo[i]; o--; ) {
int j = ej[i][o];
if (j != p) {
if ((i ^ j) == 1)
x = (x + dfs(i, j, d + 1)) % MD;
else {
int j_ = j >> 1, t = j & 1;
x = (x + dfs(i, j, d)) % MD;
x = (x - (long long) d * (min(xx[ii[j_ << 1 | 1]][t], xx[ii[i_ << 1 | 1]][s]) - max(xx[ii[j_ << 1 | 0]][t], xx[ii[i_ << 1 | 0]][s]) + 1)) % MD;
}
}
}
return x;
}
int solve() {
int h, i, j, k, l, x, y, ans;
n_ = 0, x = 0, y = 0;
for (i = 0; i < n; i++)
for (l = 0; l < ll[i]; l++)
add(x, y, x + dx[hh[i]], y + dy[hh[i]]), x += dx[hh[i]], y += dy[hh[i]];
for (i = 0; i < n_; i++)
ii[i] = i;
sort(ii, 0, n_);
n_ /= 2;
for (i = 0; i < n_ * 2; i++)
ej[i] = (int *) malloc(2 * sizeof *ej[i]), eo[i] = 0;
for (i = 0; i < n_; i++)
append(i << 1 | 0, i << 1 | 1), append(i << 1 | 1, i << 1 | 0);
for (h = -1, i = 0; i < n_; h = i, i = j) {
int y = yy[ii[i << 1 | 0]];
j = i + 1;
while (j < n_ && yy[ii[j << 1 | 0]] == y)
j++;
if (h != -1) {
k = h, l = i;
while (k < i && l < j)
if (xx[ii[k << 1 | 1]][1] < xx[ii[l << 1 | 1]][0]) {
if (xx[ii[l << 1 | 0]][0] <= xx[ii[k << 1 | 1]][1])
append(k << 1 | 1, l << 1 | 0), append(l << 1 | 0, k << 1 | 1);
k++;
} else {
if (xx[ii[k << 1 | 0]][1] <= xx[ii[l << 1 | 1]][0])
append(k << 1 | 1, l << 1 | 0), append(l << 1 | 0, k << 1 | 1);
l++;
}
}
}
ans = 0;
for (i = 0; i < n_; i++) {
int i0 = ii[i << 1 | 0], i1 = ii[i << 1 | 1];
if (yy[i0] == 0 && xx[i0][0] <= 0 && 0 <= xx[i1][0]) {
ans = dfs(-1, i << 1 | 0, 0);
for (i = 0; i < n_ * 2; i++)
free(ej[i]);
break;
}
if (yy[i1] + 1 == 0 && xx[i0][1] <= 0 && 0 <= xx[i1][1]) {
ans = dfs(-1, i << 1 | 1, 0);
for (i = 0; i < n_ * 2; i++)
free(ej[i]);
break;
}
}
return ans;
}
int draw_territory(int n_, int a, int b, vi hh_, vi ll_) {
int h, i, x, y, ans1, ans2;
long long area2, boundary, internal;
n = n_;
for (i = 0; i < n; i++)
hh[i] = hh_[i] - 1, ll[i] = ll_[i];
x = 0, y = 0, area2 = 0, boundary = 0;
for (i = 0; i < n; i++) {
h = hh[i];
area2 += cross(x, y, x + dx[h] * ll[i], y + dy[h] * ll[i]);
boundary += ll[i];
x += dx[h] * ll[i], y += dy[h] * ll[i];
}
if (area2 < 0)
area2 = -area2;
internal = (area2 - boundary) / 2 + 1;
ans1 = (boundary + internal) % MD, ans2 = 0;
if (n == 3)
ans2 = choose3(ll[0] + 2) * 2 % MD;
else if (b > 0) {
ans2 = 0;
for (h = 0; h < 3; h++) {
ans2 = (ans2 + solve()) % MD;
for (i = 0; i < n; i++)
hh[i] = (hh[i] + 1) % 6;
}
if (ans2 < 0)
ans2 += MD;
ans2 = (long long) ans2 * V2 % MD;
}
return ((long long) ans1 * a + (long long) ans2 * b) % MD;
}
Compilation message (stderr)
hexagon.cpp: In function 'void append(int, int)':
hexagon.cpp:71:23: warning: suggest parentheses around '-' in operand of '&' [-Wparentheses]
71 | if (o >= 2 && (o & o - 1) == 0)
| ~~^~~
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