-
Notifications
You must be signed in to change notification settings - Fork 0
/
Bomb Enemy.cpp
339 lines (286 loc) · 10.7 KB
/
Bomb Enemy.cpp
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
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
//
// Bomb Enemy.cpp
// LeetCode by zhaowei
//
// Created by Zhao Wei on 16/8/23.
// Copyright © 2016年 Zhao Wei. All rights reserved.
//
///@file Bomb Enemy
///@author zhao wei
///@date 2016.08.23
///@version 1.0 TLE
#include <vector>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
class Solution_v1_0 {
public:
///@brief 给定一个二维棋盘,其中E表示敌人,W表示墙,0表示空位,计算出能够杀死的最大敌人数。炸弹有且仅有一枚,只能放在空位上,其能够消灭所在行和列上的所有敌人,但是会被墙挡住
///@param grid 棋盘
///@return 返回杀死的最大敌人数
///@note 1. 枚举
// 2. 对于所有空位进行遍历,找到能够杀死的最大敌人数
// 3. 时间复杂度为O(m*n*(m + n)),空间复杂度为O(1),其中m和n分别是行列数
// 4. TLE
int maxKilledEnemies(vector<vector<char>>& grid) {
if (grid.empty() || grid[0].empty()) return 0;
int rslt = 0;
int row = grid.size(), col = grid[0].size();
for (int i = 0; i != row; i++) {
for (int j = 0; j != col; j++) {
if (grid[i][j] == 'W' || grid[i][j] == 'E') continue;
else rslt = max(rslt, killedEnemies(i, j, grid));
}
}
return rslt;
}
///@brief 计算炸弹在指定位置能够杀死的敌人数
///@param row, col 炸弹位置
///@param grid 棋盘
///@return 返回杀死的敌人数
int killedEnemies(const int& row, const int& col, vector<vector<char>>& grid) {
int kill_enemies_cnt = 0;
int l = col - 1, r = col + 1, u = row - 1, d = row + 1;
while (l >= 0) {
if (grid[row][l] == 'E') {
kill_enemies_cnt++;
l--;
}
else if (grid[row][l] == 'W') break;
else l--;
}
while (u >= 0) {
if (grid[u][col] == 'E') {
kill_enemies_cnt++;
u--;
}
else if (grid[u][col] == 'W') break;
else u--;
}
while (r < grid[0].size()) {
if (grid[row][r] == 'E') {
kill_enemies_cnt++;
r++;
}
else if (grid[row][r] == 'W') break;
else r++;
}
while (d < grid.size()) {
if (grid[d][col] == 'E') {
kill_enemies_cnt++;
d++;
}
else if (grid[d][col] == 'W') break;
else d++;
}
return kill_enemies_cnt;
}
};
class Solution_v1_1 {
struct segment {
int start_x, start_y, end_x, end_y;
int enemies;
segment(int sx, int sy, int ex, int ey, int enemies) : start_x(sx), start_y(sy), end_x(ex), end_y(ey), enemies(enemies) {}
};
vector<segment> row_segments, col_segments;
public:
///@note 1. 对棋盘进行预处理,计算出每行和每列被墙所阻隔的区间内的敌人数目;
// 2. 对所有相交区间进行遍历,找出相交区间上敌人的最大数目,交点必须为'0'。
// 3. 时间复杂度为O((m + k) * l),其中m是行数,k是墙的数目,l是区间的平均长度
// 4. TLE
int maxKilledEnemies(vector<vector<char>>& grid) {
if (grid.empty() || grid[0].empty()) return 0;
int rslt = 0;
getRowSeg(grid);
getColSeg(grid);
for (int i = 0; i != row_segments.size(); i++) {
for (int j = 0; j != col_segments.size(); j++) {
rslt = max(rslt, killEnemiesOfTwoSeg(row_segments[i], col_segments[j], grid));
}
}
return rslt;
}
///@brief 对棋盘的行区间进行预处理
///@param grid 棋盘
void getRowSeg(vector<vector<char>>& grid) {
for (int i = 0; i != grid.size(); i++) {
int wall_x = -1, wall_y = -1;
int enemies_cnt = 0;
for (int j = 0; j != grid[0].size(); j++) {
if (grid[i][j] == 'W') {
if (wall_x == -1 && wall_x == -1) {
wall_x = i;
wall_y = j;
if (j > 0) {
segment seg(i, 0, i, j-1, enemies_cnt);
row_segments.push_back(seg);
enemies_cnt = 0;
}
}
else {
segment seg(i, wall_y+1, i, j-1, enemies_cnt);
row_segments.push_back(seg);
enemies_cnt = 0;
wall_x = i;
wall_y = j;
}
}
else if (grid[i][j] == 'E') {
enemies_cnt++;
}
}
if (wall_x == -1 && wall_y == -1) {
segment seg(i, 0, i, grid[0].size()-1, enemies_cnt);
row_segments.push_back(seg);
}
else {
segment seg(i, wall_y+1, i, grid[0].size()-1, enemies_cnt);
row_segments.push_back(seg);
}
}
}
///@brief 对棋盘的列区间进行预处理
///@param grid 棋盘
void getColSeg(vector<vector<char>>& grid) {
for (int j = 0; j != grid[0].size(); j++) {
int wall_x = -1, wall_y = -1;
int enemies_cnt = 0;
for (int i = 0; i != grid.size(); i++) {
if (grid[i][j] == 'W') {
if (wall_x == -1 && wall_x == -1) {
wall_x = i;
wall_y = j;
if (i > 0) {
segment seg(0, j, i-1, j, enemies_cnt);
col_segments.push_back(seg);
enemies_cnt = 0;
}
}
else {
segment seg(wall_x+1, j, i-1, j, enemies_cnt);
col_segments.push_back(seg);
wall_x = i;
wall_y = j;
enemies_cnt = 0;
}
}
else if (grid[i][j] == 'E') {
enemies_cnt++;
}
}
if (wall_x == -1 && wall_y == -1) {
segment seg(0, j, grid.size()-1, j, enemies_cnt);
col_segments.push_back(seg);
}
else {
segment seg(wall_x+1, j, grid.size()-1, j, enemies_cnt);
col_segments.push_back(seg);
}
}
}
///@brief 判断两个区间的最大杀敌数目
///@param s1 行区间
///@param s2 列区间
///@param grid 棋盘
///@return 返回两个区间的的敌人数
int killEnemiesOfTwoSeg(segment& s1, segment& s2, vector<vector<char>>& grid) {
if (s1.start_y <= s2.start_y && s1.end_y >= s2.start_y
&& s2.start_x <= s1.start_x && s2.end_x >= s1.start_x
&& grid[s1.start_x][s2.start_y] == '0')
return s1.enemies + s2.enemies;
return 0;
}
};
class Solution_1_2 {
public:
///@note 1. 在v1.0的基础上进行了精简,在每一次遍历的过程中寻找其所能扩展的最大杀敌数。
// 2. 时间复杂度为O(mnk),空间复杂度为O(1),其中m和n分别是行列数,k为每个空格处能够扩展的平均长度。
int maxKilledEnemies(vector<vector<char>>& grid) {
if (grid.empty() || grid[0].empty()) return 0;
int row = grid.size(), col = grid[0].size();
int rslt = 0;
for (int i = 0; i != row; i++) {
for (int j = 0; j != col; j++) {
int sum = 0;
if (grid[i][j] == '0') {
int up = i - 1, down = i + 1, left = j - 1, right = j + 1;
while (up >= 0 && grid[up][j] != 'W') {
if (grid[up][j] == 'E') sum++;
up--;
}
while (down < row && grid[down][j] != 'W') {
if (grid[down][j] == 'E') sum++;
down++;
}
while (left >= 0 && grid[i][left] != 'W') {
if (grid[i][left] == 'E') sum++;
left--;
}
while (right < col && grid[i][right] != 'W') {
if (grid[i][right] == 'E') sum++;
right++;
}
rslt = sum > rslt ? sum : rslt;
}
}
}
return rslt;
}
};
class Solution {
public:
///@note 1. 设置一个临时变量来保存每一行中当前连续的敌人数;
// 2. 设置一个临时数组保存每一列的当前连续敌人数;
// 3. 在遍历的时候因为在之前已经计算过当前位置的行列连续敌人数,直接相加即可;
// 4. 时间复杂度为O(mn),空间复杂度为O(n),其中m是行数,n是列数。
int maxKilledEnemies(vector<vector<char>>& grid) {
if (grid.empty() || grid[0].empty()) return 0;
int row = grid.size(), col = grid[0].size();
int rslt = 0;
int row_hit = 0;
vector<int> col_hit(col, 0);
for (int i = 0; i != row; i++) {
for (int j = 0; j != col; j++) {
if (!j || grid[i][j-1] == 'W') {
row_hit = 0;
for (int k = j; k < col && grid[i][k] != 'W'; k++) {
row_hit += grid[i][k] == 'E';
}
}
if (!i || grid[i-1][j] == 'W') {
col_hit[j] = 0;
for (int k = i; k < row && grid[k][j] != 'W'; k++) {
col_hit[j] += grid[k][j] == 'E';
}
}
if (grid[i][j] == '0')
rslt = max(rslt, row_hit + col_hit[j]);
}
}
return rslt;
}
};
int main() {
/*
0 E 0 0
E 0 W E
0 E 0 0
*/
vector<vector<char>> grid;
vector<char> line;
line.push_back('W');
line.push_back('E');
line.push_back('W');
line.push_back('0');
line.push_back('E');
grid.push_back(line);
line.clear();
Solution_v1_0 slt_v1_0;
int rslt = slt_v1_0.maxKilledEnemies(grid);
Solution_v1_1 slt_v1_1;
rslt = slt_v1_1.maxKilledEnemies(grid);
Solution slt;
rslt = slt.maxKilledEnemies(grid);
return 0;
}