问题描述：
一个城市被表示为一个二维矩形矩阵。给定矩阵的外墙表示城市的边界。市民分散在城市的不同地点。它们要么用{a, c} 表示。冠状病毒已经感染了这座城市。
冠状病毒从坐标(0, 0)进入城市并沿对角线路径传播，直到遇到人类。如果它遇到一个被指定为的人，它的轨迹会逆时针（从右到左）旋转90 度。类似地，如果它遇到一个被指定为c 的人，它的轨迹会顺时针（从左到右）旋转90 度。感染人后，病毒继续沿其对角线移动。
在它的穿越过程中，如果它第一次到达城市边界，它会旋转90度重新进入城市。然而，如果它第二次击中任何边界墙，病毒就会被摧毁。
你必须计算病毒的轨迹，打印可以找到受感染公民的城市地图，最后报告安全和受感染公民的数量。
输入：
一个9 行20 列的输入矩阵，由字符“ *” 、 “a” 、 “c”和“.”组成。在哪里
- “*”表示城市边界上的元素
- “a”表示遇到病毒轨迹改变90度（逆时针方向）的公民
- “c”表示遇到病毒轨迹改变90度（顺时针方向）的公民
- “。” （点）表示城市内的空位
输出：
随机数的行，每行表示病毒轨迹的坐标。
从下一行开始，一个9 行20 列的输出矩阵由字符“ *” 、 “a” 、 “c” 、 “.”组成。和“-”在哪里
- “*”表示城市边界上的元素
- “a”表示遇到病毒轨迹改变90度（逆时针方向）的公民
- “c”表示遇到病毒轨迹改变90度（顺时针方向）的公民
- “。” （点）表示城市内的空位
- “-”表示受感染公民所在的位置
接下来的两行打印了城市中安全和受感染公民的数量。
约束：
- 0 <= x <= 20
- 0 <= y <= 8
- 病毒打不到三个角 (20, 8) (20, 0) (0, 8)
例子：
Input:
********************
*….c………….*
*…c…………..*
*c……………..*
*………….a….*
*c.c……………*
*.a…………….*
*………..c……*
********************
Output:
0 0
1 1
2 2
1 3
2 4
3 5
4 6
5 5
6 4
7 3
8 2
9 1
10 0
11 1
12 2
13 3
14 4
13 5
12 6
11 7
10 8
********************
*….c………….*
*…-…………..*
*c……………..*
*………….-….*
*-.c……………*
*.-…………….*
*………..c……*
********************
safe=4
infected=4
Explanation: The virus trajectory starts from (0, 0) and crosses (1, 1) (2, 2). At (2, 2) we have a citizen of type a causing the virus trajectory to be rotated by 90 degrees anti-clockwise. It moves until the next citizen in its path is encountered at (1, 3) of type c causing the virus trajectory to be rotated by 90 degree clockwise. It continues on its path till it reaches the next human in its path at (4, 6) of type c Virus trajectory is again rotated by 90 degrees clockwise until it hits the boundary at (10, 0). Since this is the first time that the virus hits the boundary, it rotates by 90 degrees anticlockwise to reenter the city. The trajectory then continues towards (11, 1) (12, 2) (13, 3) and finally a citizen at (14, 4) of type a rotating the trajectory to 90 degree anticlockwise. From there it continues its trajectory and hits the boundary at (10, 8).
Since this is the second time the virus hits the boundary, the virus is destroyed.
So, along its trajectory starting from (0, 0) it has infected 4 citizens at location (2, 2) (1, 3) (4, 6) (14, 4). The other 4 citizens who did not come in the virus trajectory are deemed to be safe.
Input:
********************
*………………*
*..c……………*
*….c………….*
*………a……..*
*………………*
*…….a……c…*
*………………*
********************
Output:
0 0
1 1
2 2
3 3
4 4
5 5
6 4
7 3
8 2
9 3
10 4
9 5
8 6
7 7
6 8
5 7
4 6
3 5
2 4
1 3
0 2
********************
*………………*
*..c……………*
*….-………….*
*………-……..*
*………………*
*…….-……c…*
*………………*
********************
safe=2
infected=3
Explanation: The virus trajectory starts from (0, 0) and crosses (1, 1) (2, 2) (5, 5). At (5, 5) we have a citizen of type c causing the virus trajectory to be rotated by 90 degrees clockwise. It moves until the next citizen in its path is encountered at (8, 2) of type a causing the virus trajectory to be rotated by 90 degree anti-clockwise. It continues on its path till it reaches the next human in its path at (10, 4) of type a Virus trajectory is again rotated by 90 degrees clockwise until it hits the boundary at (6, 8). Since this is the first time that the virus hits the boundary, it rotates by 90 degrees anticlockwise to reenter the city. The trajectory then continues towards (9, 5) (8, 6) (7, 7) (6, 8) and renters the city by rotating the trajectory by 90 degrees anti-clockwise to follow the trajectory (5, 7) (4, 6) (3, 5) (2, 4) (1, 3) (0, 2).
At (0, 2) it again hits the boundary and since this is the second time the virus hits the boundary, the virus is destroyed. So along its trajectory starting from (0, 0) it has infected 3 citizens at location (5, 5) (10, 4) (8, 2). The other 2 citizens who did not come in the virus trajectory are deemed to be safe.
方法：解决这个问题的思路是先将输入所表示的图求反，根据给定的条件遍历图。请按照以下步骤解决问题：
- 首先，将输入数组反转为反转数组。所以最后一行变成第一行，倒数第二行变成第二行，依此类推。这样做是为了在具有正x和y的二维坐标系中查看图形。
- 遍历图形并在名为bound的变量中跟踪病毒到达边界的次数并运行 while 循环，直到该变量的值小于或等于1 。
- 初始化一个变量direct ，将运动方向保持为：
- 如果 direct 为 1，则方向为东北。
- 如果 direct 为 2，则方向为西北。
- 如果 direct 为 3，则方向为东南。
- 如果 direct 为 4，则方向为西南。
- 病毒将有 4 个运动方向。病毒将从顶点(0, 0)开始，运动方向为1 （向东北方向移动）。当遇到墙壁时，变量的方向会发生变化。
- 此外，当遇到‘a’时，将direct的值更改为产生的逆时针方向，当遇到‘c’时，将direct的值更改为从当前方向顺时针旋转90 度。
- 对于‘.’ （点）字符，保持当前位置变量递增或递减1以在直接变量的描述方向上对角移动。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
#define row 9
#define col 20
// Funtion to print trajectory of the virus
// and safe and infected citizens
void coronaVirus(char arr[row][col])
{
// To store Inverted Array
char inverted[9][20];
// Temporary array to store
// original array
char temp[9][20];
// To store number of infected citizens
int count = 0;
// To store total number of citizens ('a', 'c')
int total = 0;
// To invert the array
for (int i = 0; i <= 8; i++) {
for (int j = 0; j < 20; j++) {
// Invert the array row wise
inverted[i][j] = arr[8 - i][j];
}
}
// Count the number of citiens
for (int i = 0; i <= 8; i++) {
for (int j = 0; j < 20; j++) {
// Count total number of citizens.
if (inverted[i][j] == 'a'
|| inverted[i][j] == 'c') {
total++;
}
// Copy inverted array in temp
temp[i][j] = inverted[i][j];
}
}
// To store number of times,
// virus encountered a boundary
int bound = 0;
// Variable for row-wise traversal
int i = 1;
// Variable for column-wise traversal
int j = 1;
// Variable to store direction
int direct = 1;
// The virus starts from (0, 0) always
cout << "0 0" << endl;
// To break infinite looping
int flag = 0;
// Finding trajectory and safe
// and infected citizens
while (bound <= 1 && flag < 1000) {
flag++;
cout << i << " " << j << endl;
// Virus has striked the boundary twice.
// Therefore, end loop
if (inverted[j][i] == '*' && bound == 1) {
break;
}
// Virus striked the corners, end loop
if (inverted[j][i] == '*') {
if (i == 0 && j == 8 || i == 20 && j == 8
|| i == 20 && j == 0) {
break;
}
}
// First time for the virus
// to hit the boundary
if (inverted[j][i] == '*' && bound == 0) {
// Increment bound variable
bound++;
// Srikes the left wall
if (i == 0 && j != 8 && j != 0) {
// Turns 90 degree clockwise
if (direct == 2) {
direct = 1;
i++;
j++;
}
// Else turns 90 degree anticlockwise
else if (direct == 4) {
direct = 3;
j--;
i++;
}
}
// Strikes the right wall
else if (i == 20 && j != 0 && j != 8) {
// Turns 90 degree anticlockwise
if (direct == 1) {
direct = 2;
i--;
j++;
}
// Else turns 90 degree clockwise
else if (direct == 3) {
direct = 4;
i--;
j--;
}
}
// Strikes the upper wall
else if (j == 8 && i != 0 && i != 20) {
// Turns 90 degree anticlockwise
if (direct == 2) {
direct = 4;
i--;
j--;
}
// Else turns 90 degree clockwise
else if (direct == 1) {
direct = 3;
j--;
i++;
}
}
// Strikes the lower wall
else if (j == 0 && i != 0 && i != 20) {
// Turns 90 degree clockwise
if (direct == 4) {
direct = 2;
i--;
j++;
}
// Else turns 90 degree anticlockwise
else if (direct == 3) {
direct = 1;
i++;
j++;
}
}
continue;
}
// Make 'c' visited by replacing it by'-'
if (inverted[j][i] == 'c') {
temp[j][i] = '-';
// Turns all directions 90
// degree clockwise
if (direct == 1) {
direct = 3;
j--;
i++;
}
else if (direct == 2) {
direct = 1;
i++;
j++;
}
else if (direct == 3) {
direct = 4;
i--;
j--;
}
else if (direct == 4) {
direct = 2;
i--;
j++;
}
// Increment count of infected citizens
count++;
continue;
}
// Make 'a' visited by replacing it by'-'
if (inverted[j][i] == 'a') {
temp[j][i] = '-';
// Turns all directions by 90 degree
// anticlockwise
if (direct == 1) {
direct = 2;
i--;
j++;
}
else if (direct == 2) {
direct = 4;
i--;
j--;
}
else if (direct == 3) {
direct = 1;
i++;
j++;
}
else if (direct == 4) {
direct = 3;
j--;
i++;
}
// Increment count of
// infected citizens
count++;
continue;
}
// Increment the counter diagonally
// in the given direction.
if (inverted[j][i] == '.') {
if (direct == 1) {
i++;
j++;
}
else if (direct == 2) {
i--;
j++;
}
else if (direct == 3) {
j--;
i++;
}
else if (direct == 4) {
i--;
j--;
}
continue;
}
}
// Print the mirror of the array
// i.e. last row must be printed first.
for (int i = 0; i <= 8; i++) {
for (int j = 0; j < 20; j++) {
cout << temp[8 - i][j];
}
cout << endl;
}
// Print safe and infected citizens
cout << "safe=" << (total - count) << endl;
cout << "infected=" << (count) << endl;
}
// Driver Code
int main()
{
// Given 2D array
char arr[row][col]
= { { '*', '*', '*', '*', '*', '*', '*',
'*', '*', '*', '*', '*', '*', '*',
'*', '*', '*', '*', '*', '*' },
{ '*', '.', '.', '.', '.', '.', '.',
'.', '.', '.', '.', '.', '.', '.',
'.', '.', '.', '.', '.', '*' },
{ '*', '.', '.', 'c', '.', '.', '.',
'.', '.', '.', '.', '.', '.', '.',
'.', '.', '.', '.', '.', '*' },
{ '*', '.', '.', '.', '.', 'c', '.',
'.', '.', '.', '.', '.', '.', '.',
'.', '.', '.', '.', '.', '*' },
{ '*', '.', '.', '.', '.', '.', '.',
'.', '.', '.', 'a', '.', '.', '.',
'.', '.', '.', '.', '.', '*' },
{ '*', '.', '.', '.', '.', '.', '.',
'.', '.', '.', '.', '.', '.', '.',
'.', '.', '.', '.', '.', '*' },
{ '*', '.', '.', '.', '.', '.', '.',
'.', 'a', '.', '.', '.', '.', '.',
'.', 'c', '.', '.', '.', '*' },
{ '*', '.', '.', '.', '.', '.', '.',
'.', '.', '.', '.', '.', '.', '.',
'.', '.', '.', '.', '.', '*' },
{ '*', '*', '*', '*', '*', '*', '*',
'*', '*', '*', '*', '*', '*', '*',
'*', '*', '*', '*', '*', '*' } };
// Function Call
coronaVirus(arr);
return 0;
} 0 0
1 1
2 2
3 3
4 4
5 5
6 4
7 3
8 2
9 3
10 4
9 5
8 6
7 7
6 8
5 7
4 6
3 5
2 4
1 3
0 2
********************
*..................*
*..c...............*
*....-.............*
*.........-........*
*..................*
*.......-......c...*
*..................*
********************
safe=2
infected=3
时间复杂度： O(R*C) 其中 R 是行数，C 是列数
辅助空间： O(R*C)
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