#include <cstdio> #include <unordered_map> #include <string> #include <vector> #include <cassert> #define LCM (8 * 7 * 5 * 3) #define BML (LCM / 64 + 1) #define MAXN 16000 typedef long long ll; using std::string; typedef struct bitmask { ll mask[BML]; void clean() { for (int i = 0; i < BML; i++) mask[i] = 0LL; } void fill() { for (int i = 0; i < BML; i++) mask[i] = -1LL; } void set_at(int pos) { int base = pos / 64; int off = pos & 63; mask[base] |= (1LL << off); } bool get(int pos) { int base = pos / 64; int off = pos & 63; return mask[base] & (1LL << off); } string debug(int L) { string res = ""; for (int i = 0; i < L; ++i) { res += (get(i) ? '1' : '0'); } return res; } void do_or(const bitmask &b) { for (int i = 0; i < BML; ++i) mask[i] |= b.mask[i]; } void do_and(const bitmask &b) { for (int i = 0; i < BML; ++i) mask[i] &= b.mask[i]; } void do_not() { for (int i = 0; i < BML; ++i) mask[i] = ~mask[i]; } } bitmask; std::unordered_map<string, bitmask> string_to_bitmask; void fill_string_to_bitmask() { for (int l = 2; l < 9; ++l) { for (int code = 0; code < (1 << l); ++code) { std::string s = ""; s.resize(l, '0'); bitmask b; b.clean(); for (int p = 0; p < l; ++p) { if (code & (1 << p)) { s[p] = '1'; b.set_at(p); } } for (int p = l; p < LCM; ++p) { if (b.get(p - l)) b.set_at(p); } string_to_bitmask[s] = b; } // for (int code) } // for(int l) } std::vector<std::vector<bitmask>> intersections; int N, M, Q; char tempstring[20]; bitmask vertical_masks[MAXN]; // Length M, 1 if I can cross the vertical street from i to i+1 bitmask horizontal_masks[MAXN]; // Length N, 1 if I can cross the horizontal street from i to i+1. // vertical_forward_times[time_offset][jump_size][m] = X means it takes X time to get from plaza // (2 ** jump_size * m to 2**jump_size * (m + 1), if we start at time time_offset. std::vector<int> vertical_forward_times[LCM][16]; // vertical_backward_times measure the time to get from 2 ** jumpsize * (m+1) to 2 ** jumpsize * m std::vector<int> vertical_backward_times[LCM][16]; // horizontal_forward_times do the same from n to n+1 std::vector<int> horizontal_forward_times[LCM][16]; std::vector<int> horizontal_backward_times[LCM][16]; int calculate(int time, int from, int to, std::vector<int> forward[LCM][16], std::vector<int> backward[LCM][16]) { int cost = 0; if (from <= to) { while (from < to) { // Get the jump length. int jumplen = 0; while (from % (1 << (jumplen + 1)) == 0 && from + (1 << (jumplen + 1)) <= to) { jumplen += 1; } // Move by jumplen. int movecost = forward[time][jumplen][from / (1 << jumplen)]; from += (1 << jumplen); cost += movecost; time = (time + movecost) % LCM; } return cost; } // The reverse, let's get this right. while (from > to) { int jumplen = 0; while (from % (1 << (jumplen + 1)) == 0 && from - (1 << (jumplen + 1)) >= to) { jumplen += 1; } int movecost = backward[time][jumplen][(from / (1 << jumplen)) - 1]; from -= (1 << jumplen); cost += movecost; time = (time + movecost) % LCM; } return cost; } int main() { // Fill in the string to bitmask converstion. fill_string_to_bitmask(); // Set up the bitmasks on every intersection. scanf("%d %d %d\n", &N, &M, &Q); intersections.resize(N); for (int n = 0; n < N; ++n) { intersections[n].resize(M); for (int m = 0; m < M; ++m) { scanf("%s", tempstring); intersections[n][m] = string_to_bitmask[string(tempstring)]; } } // Calculate vertical masks - is there a crossing between vertical X and X+1. // There is no crossing if all the lights are on 0. Otherwise, there is. So, we OR the bitmasks. for (int m = 0; m < M; ++m) { vertical_masks[m].clean(); for (int n = 0; n < N; ++n) { vertical_masks[m].do_or(intersections[n][m]); } // Now, it's a 1 if we can cross (if any light allowed a horizontal crossing). } for (int n = 0; n < N; ++n) { horizontal_masks[n].fill(); for (int m = 0; m < M; ++m) { horizontal_masks[n].do_and(intersections[n][m]); } // Now, it's a 0 if we can cross (if any light allowed a vertical crossing). Let's reverse it. horizontal_masks[n].do_not(); } // Let's calculate times. First, resize the vectors. for (int time_offset = 0; time_offset < LCM; ++time_offset) { for (int jump_size = 0; jump_size < 16; ++jump_size) { vertical_forward_times[time_offset][jump_size].resize(M / (1 << jump_size)); vertical_backward_times[time_offset][jump_size].resize(M / (1 << jump_size)); horizontal_forward_times[time_offset][jump_size].resize(N / (1 << jump_size)); horizontal_backward_times[time_offset][jump_size].resize(N / (1 << jump_size)); } } // Begin with time offset 0. // Now, I want the nearest zero, moving across time offsets forwards in time. // The last one, let's just do by hand. for (int m = 0; m < M; ++m) { int diff = 0; while (!vertical_masks[m].get((LCM + diff - 1) % LCM)) diff++; // Note that at jump size zero, the forward and backward times are equal - we just need to cross one crossing. vertical_forward_times[LCM-1][0][m] = vertical_backward_times[LCM-1][0][m] = diff; // The rest, we do with the standard DP. for (int time_offset = LCM - 2; time_offset >= 0; time_offset--) { vertical_forward_times[time_offset][0][m] = vertical_backward_times[time_offset][0][m] = (vertical_masks[m].get(time_offset) ? 0 : vertical_forward_times[time_offset + 1][0][m] + 1); } } // Similarly for horizontals. for (int n = 0; n < N; ++n) { int diff = 0; while (!horizontal_masks[n].get((LCM + diff - 1) % LCM)) diff++; horizontal_forward_times[LCM-1][0][n] = horizontal_backward_times[LCM-1][0][n] = diff; for (int time_offset = LCM - 2; time_offset >= 0; time_offset--) { horizontal_forward_times[time_offset][0][n] = horizontal_backward_times[time_offset][0][n] = (horizontal_masks[n].get(time_offset) ? 0 : horizontal_forward_times[time_offset + 1][0][n] + 1); } } // Now, do the jumps. They are a simple aggregation. for (int jump_size = 1; jump_size < 16; ++jump_size) { for (int m = 0; (m + 1) * (1 << jump_size) < M + 1; ++m) { for (int time_offset = 0; time_offset < LCM; ++time_offset) { int first_jump = vertical_forward_times[time_offset][jump_size - 1][2 * m]; int second_time = (time_offset + first_jump) % LCM; vertical_forward_times[time_offset][jump_size][m] = first_jump + vertical_forward_times[second_time][jump_size - 1][2 * m + 1]; first_jump = vertical_backward_times[time_offset][jump_size - 1][2 * m + 1]; second_time = (time_offset + first_jump) % LCM; vertical_backward_times[time_offset][jump_size][m] = first_jump + vertical_backward_times[second_time][jump_size - 1][2 * m]; } } for (int n = 0; (n + 1) * (1 << jump_size) < N + 1; ++n) { for (int time_offset = 0; time_offset < LCM; ++time_offset) { int first_jump = horizontal_forward_times[time_offset][jump_size - 1][2 * n]; int second_time = (time_offset + first_jump) % LCM; horizontal_forward_times[time_offset][jump_size][n] = first_jump + horizontal_forward_times[second_time][jump_size - 1][2 * n + 1]; first_jump = horizontal_backward_times[time_offset][jump_size - 1][2 * n + 1]; second_time = (time_offset + first_jump) % LCM; horizontal_backward_times[time_offset][jump_size][n] = first_jump + horizontal_backward_times[second_time][jump_size - 1][2 * n]; } } } // Aaaah, this was tiring. OK, let's do the queries. for (int q = 0; q < Q; ++q) { int starttime, nstart, mstart, nend, mend; scanf("%d %d %d %d %d", &starttime, &nstart, &mstart, &nend, &mend); // Time to cross the horizontal streets. int htime = calculate(starttime % LCM, nstart, nend, horizontal_forward_times, horizontal_backward_times); int vtime = calculate(starttime % LCM, mstart, mend, vertical_forward_times, vertical_backward_times); assert(htime == 0 || vtime == 0); printf("%d\n", starttime + htime + vtime); } return 0; }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 | #include <cstdio> #include <unordered_map> #include <string> #include <vector> #include <cassert> #define LCM (8 * 7 * 5 * 3) #define BML (LCM / 64 + 1) #define MAXN 16000 typedef long long ll; using std::string; typedef struct bitmask { ll mask[BML]; void clean() { for (int i = 0; i < BML; i++) mask[i] = 0LL; } void fill() { for (int i = 0; i < BML; i++) mask[i] = -1LL; } void set_at(int pos) { int base = pos / 64; int off = pos & 63; mask[base] |= (1LL << off); } bool get(int pos) { int base = pos / 64; int off = pos & 63; return mask[base] & (1LL << off); } string debug(int L) { string res = ""; for (int i = 0; i < L; ++i) { res += (get(i) ? '1' : '0'); } return res; } void do_or(const bitmask &b) { for (int i = 0; i < BML; ++i) mask[i] |= b.mask[i]; } void do_and(const bitmask &b) { for (int i = 0; i < BML; ++i) mask[i] &= b.mask[i]; } void do_not() { for (int i = 0; i < BML; ++i) mask[i] = ~mask[i]; } } bitmask; std::unordered_map<string, bitmask> string_to_bitmask; void fill_string_to_bitmask() { for (int l = 2; l < 9; ++l) { for (int code = 0; code < (1 << l); ++code) { std::string s = ""; s.resize(l, '0'); bitmask b; b.clean(); for (int p = 0; p < l; ++p) { if (code & (1 << p)) { s[p] = '1'; b.set_at(p); } } for (int p = l; p < LCM; ++p) { if (b.get(p - l)) b.set_at(p); } string_to_bitmask[s] = b; } // for (int code) } // for(int l) } std::vector<std::vector<bitmask>> intersections; int N, M, Q; char tempstring[20]; bitmask vertical_masks[MAXN]; // Length M, 1 if I can cross the vertical street from i to i+1 bitmask horizontal_masks[MAXN]; // Length N, 1 if I can cross the horizontal street from i to i+1. // vertical_forward_times[time_offset][jump_size][m] = X means it takes X time to get from plaza // (2 ** jump_size * m to 2**jump_size * (m + 1), if we start at time time_offset. std::vector<int> vertical_forward_times[LCM][16]; // vertical_backward_times measure the time to get from 2 ** jumpsize * (m+1) to 2 ** jumpsize * m std::vector<int> vertical_backward_times[LCM][16]; // horizontal_forward_times do the same from n to n+1 std::vector<int> horizontal_forward_times[LCM][16]; std::vector<int> horizontal_backward_times[LCM][16]; int calculate(int time, int from, int to, std::vector<int> forward[LCM][16], std::vector<int> backward[LCM][16]) { int cost = 0; if (from <= to) { while (from < to) { // Get the jump length. int jumplen = 0; while (from % (1 << (jumplen + 1)) == 0 && from + (1 << (jumplen + 1)) <= to) { jumplen += 1; } // Move by jumplen. int movecost = forward[time][jumplen][from / (1 << jumplen)]; from += (1 << jumplen); cost += movecost; time = (time + movecost) % LCM; } return cost; } // The reverse, let's get this right. while (from > to) { int jumplen = 0; while (from % (1 << (jumplen + 1)) == 0 && from - (1 << (jumplen + 1)) >= to) { jumplen += 1; } int movecost = backward[time][jumplen][(from / (1 << jumplen)) - 1]; from -= (1 << jumplen); cost += movecost; time = (time + movecost) % LCM; } return cost; } int main() { // Fill in the string to bitmask converstion. fill_string_to_bitmask(); // Set up the bitmasks on every intersection. scanf("%d %d %d\n", &N, &M, &Q); intersections.resize(N); for (int n = 0; n < N; ++n) { intersections[n].resize(M); for (int m = 0; m < M; ++m) { scanf("%s", tempstring); intersections[n][m] = string_to_bitmask[string(tempstring)]; } } // Calculate vertical masks - is there a crossing between vertical X and X+1. // There is no crossing if all the lights are on 0. Otherwise, there is. So, we OR the bitmasks. for (int m = 0; m < M; ++m) { vertical_masks[m].clean(); for (int n = 0; n < N; ++n) { vertical_masks[m].do_or(intersections[n][m]); } // Now, it's a 1 if we can cross (if any light allowed a horizontal crossing). } for (int n = 0; n < N; ++n) { horizontal_masks[n].fill(); for (int m = 0; m < M; ++m) { horizontal_masks[n].do_and(intersections[n][m]); } // Now, it's a 0 if we can cross (if any light allowed a vertical crossing). Let's reverse it. horizontal_masks[n].do_not(); } // Let's calculate times. First, resize the vectors. for (int time_offset = 0; time_offset < LCM; ++time_offset) { for (int jump_size = 0; jump_size < 16; ++jump_size) { vertical_forward_times[time_offset][jump_size].resize(M / (1 << jump_size)); vertical_backward_times[time_offset][jump_size].resize(M / (1 << jump_size)); horizontal_forward_times[time_offset][jump_size].resize(N / (1 << jump_size)); horizontal_backward_times[time_offset][jump_size].resize(N / (1 << jump_size)); } } // Begin with time offset 0. // Now, I want the nearest zero, moving across time offsets forwards in time. // The last one, let's just do by hand. for (int m = 0; m < M; ++m) { int diff = 0; while (!vertical_masks[m].get((LCM + diff - 1) % LCM)) diff++; // Note that at jump size zero, the forward and backward times are equal - we just need to cross one crossing. vertical_forward_times[LCM-1][0][m] = vertical_backward_times[LCM-1][0][m] = diff; // The rest, we do with the standard DP. for (int time_offset = LCM - 2; time_offset >= 0; time_offset--) { vertical_forward_times[time_offset][0][m] = vertical_backward_times[time_offset][0][m] = (vertical_masks[m].get(time_offset) ? 0 : vertical_forward_times[time_offset + 1][0][m] + 1); } } // Similarly for horizontals. for (int n = 0; n < N; ++n) { int diff = 0; while (!horizontal_masks[n].get((LCM + diff - 1) % LCM)) diff++; horizontal_forward_times[LCM-1][0][n] = horizontal_backward_times[LCM-1][0][n] = diff; for (int time_offset = LCM - 2; time_offset >= 0; time_offset--) { horizontal_forward_times[time_offset][0][n] = horizontal_backward_times[time_offset][0][n] = (horizontal_masks[n].get(time_offset) ? 0 : horizontal_forward_times[time_offset + 1][0][n] + 1); } } // Now, do the jumps. They are a simple aggregation. for (int jump_size = 1; jump_size < 16; ++jump_size) { for (int m = 0; (m + 1) * (1 << jump_size) < M + 1; ++m) { for (int time_offset = 0; time_offset < LCM; ++time_offset) { int first_jump = vertical_forward_times[time_offset][jump_size - 1][2 * m]; int second_time = (time_offset + first_jump) % LCM; vertical_forward_times[time_offset][jump_size][m] = first_jump + vertical_forward_times[second_time][jump_size - 1][2 * m + 1]; first_jump = vertical_backward_times[time_offset][jump_size - 1][2 * m + 1]; second_time = (time_offset + first_jump) % LCM; vertical_backward_times[time_offset][jump_size][m] = first_jump + vertical_backward_times[second_time][jump_size - 1][2 * m]; } } for (int n = 0; (n + 1) * (1 << jump_size) < N + 1; ++n) { for (int time_offset = 0; time_offset < LCM; ++time_offset) { int first_jump = horizontal_forward_times[time_offset][jump_size - 1][2 * n]; int second_time = (time_offset + first_jump) % LCM; horizontal_forward_times[time_offset][jump_size][n] = first_jump + horizontal_forward_times[second_time][jump_size - 1][2 * n + 1]; first_jump = horizontal_backward_times[time_offset][jump_size - 1][2 * n + 1]; second_time = (time_offset + first_jump) % LCM; horizontal_backward_times[time_offset][jump_size][n] = first_jump + horizontal_backward_times[second_time][jump_size - 1][2 * n]; } } } // Aaaah, this was tiring. OK, let's do the queries. for (int q = 0; q < Q; ++q) { int starttime, nstart, mstart, nend, mend; scanf("%d %d %d %d %d", &starttime, &nstart, &mstart, &nend, &mend); // Time to cross the horizontal streets. int htime = calculate(starttime % LCM, nstart, nend, horizontal_forward_times, horizontal_backward_times); int vtime = calculate(starttime % LCM, mstart, mend, vertical_forward_times, vertical_backward_times); assert(htime == 0 || vtime == 0); printf("%d\n", starttime + htime + vtime); } return 0; } |