# 【題解】ZeroJudge g596: 2. 動線安排

【題目敘述】https://zerojudge.tw/ShowProblem?problemid=g596
【解題想法】模擬

```#include <iostream>
#include <map>
using namespace std;
bool myDebug = false;
const int maxn = 105;
int a[maxn][maxn]; //紀錄有幾條線經過（-1/0/1/2），-1表示有木樁
int m, n, h;
int dx[] = {-1, 1, 0, 0}; // 上(0)下(1)左(2)右(3)
int dy[] = {0, 0, -1, 1}; // 上(0)下(1)左(2)右(3)

void myPrint() {
map <int, char> mp;
mp[-1] = '@';
mp[0] = '_';
mp[1] = '*';
mp[2] = '+';
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
cout << mp[a[i][j]];
}
cout << "\n";
}
cout << "\n";
}

int main() {
cin >> m >> n >> h;
int r, c, t;
int ans = 0; //最後有線和有木樁佔據空間的面積
int mx = 0; //操作過程中有線和有木樁佔據空間的面積最大值
if (myDebug) myPrint();
for (int k = 0; k < h; k++) {
cin >> r >> c >> t;
//有線經過則先將那些線拆掉
for (int i = 0; i < 4; i++) {
int nx = r + dx[i];
int ny = c + dy[i];
bool find = false;
while (nx >= 0 && nx < m && ny >=0 && ny < n) {
if (a[nx][ny] == 0) {
//沒有線經過
break;
}

if (a[nx][ny] == -1) {
//確定有端點木樁，才不會誤拆了屬於別的木樁的線
find = true;
break;
}

nx += dx[i];
ny += dy[i];
}
if (find) {
for (int j = min(r, nx)+1; j < max(r, nx); j++) {
a[j][ny]--;
}
for (int j = min(c, ny)+1; j < max(c, ny); j++) {
a[nx][j]--;
}
}
}

if (t == 0) {
//加一木樁，然後連線
a[r][c] = -1;
for (int i = 0; i < 4; i++) {
int nx = r + dx[i];
int ny = c + dy[i];
bool find = false;
while (nx >= 0 && nx < m && ny >=0 && ny < n) {
if (a[nx][ny] == -1) {
find = true;
break;
}
nx += dx[i];
ny += dy[i];
}
if (find) {
for (int j = min(r, nx)+1; j < max(r, nx); j++) {
a[j][ny]++;
}
for (int j = min(c, ny)+1; j < max(c, ny); j++) {
a[nx][j]++;
}
}
}
} else {
//移除木樁
a[r][c] = 0;
}
ans = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
ans += (a[i][j] != 0);
}
}
mx = max(mx, ans);
if (myDebug) myPrint();
}
cout << mx << "\n" << ans << "\n";
return 0;
}

```