题意:有一个地图,有空地和墙,空地是一个连通块,问要再添k堵墙,使得空地还是一个连通块,输出改变后的地图。
解法:这题的思维很新颖,没想到啊,如果正面搜索,需要考虑的情况比较多,所以逆向思维,先把所有’.‘变成’X’,然后,再找出一个大小为ans-k数量的连通块,记得dfs的时候计数器要定义为全局变量,害,下次我还写bfs。
代码:
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define _for(n,m,i) for (register int i = (n); i < (m); ++i)
#define _rep(n,m,i) for (register int i = (n); i <= (m); ++i)
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
#define PI acos(-1)
#define eps 1e-9
#define rint register int
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)
using namespace std;
typedef long long LL;
typedef pair<LL, int> pli;
typedef pair<int, int> pii;
typedef pair<double, int> pdi;
typedef pair<LL, LL> pll;
typedef pair<double, double> pdd;
typedef map<int, int> mii;
typedef map<char, int> mci;
typedef map<string, int> msi;
template<class T>
void read(T &res) {
int f = 1; res = 0;
char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1; c = getchar(); }
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0'; c = getchar(); }
res *= f;
}
const int ne[8][2] = {
1, 0, -1, 0, 0, 1, 0, -1, -1, -1, -1, 1, 1, -1, 1, 1};
const int INF = 0x3f3f3f3f;
const int N = 510;
const LL Mod = 1e9+7;
const int M = 1e6+10;
char tu[N][N];
int n, m, k, ans, cnt;
void dfs(int x, int y) {
//cout << x << " " << y << " " << cnt << " " << ans-k << endl;
for(int i = 0, tx, ty; i < 4; ++i) {
tx = x + ne[i][0], ty = y + ne[i][1];
if(tx >=1 && ty >= 1 && tx <= n && ty <= m && tu[tx][ty] == 'X') {
if(cnt == ans-k) return;
tu[tx][ty] = '.'; ++cnt;
dfs(tx, ty);
}
}
}
int main() {
scanf("%d %d %d", &n, &m, &k);
int fx, fy;
_rep(1, n, i) {
scanf("%s", tu[i]+1);
_rep(1, m, j) if(tu[i][j] == '.') {
tu[i][j] = 'X', ++ans;
fx = i; fy = j;
}
}
tu[fx][fy] = '.'; ++cnt; dfs(fx, fy);
_rep(1, n, i) printf("%s\n", tu[i]+1);
return 0;
}