HDUOJ 1045 Fire Net

#include <iostream>#include <string>#include <vector>#include <stack>#include <queue>#include <deque>#include <set>#include <map>#include <algorithm>#include <functional>#include <sstream>#include <utility>#include <cstring>#include <cstdio>#include <cstdlib>#include <ctime>#include <cmath>#include <cctype>#define CLEAR(a, b) memset(a, b, sizeof(a))#define IN() freopen("in.txt", "r", stdin)#define OUT() freopen("out.txt", "w", stdout)#define PF(a) printf("%d\n", a)#define SF(a) scanf("%d", &a)#define SFF(a, b) scanf("%d%d", &a, &b)#define FOR(i, a, b) for(int i = a; i < b; ++i)#define LL long long#define maxn 10#define mod 1000007#define INF 1000005using namespace std;//------------------------------------------------------------------------------------------////其实想法上倒也没错啦，，就是枚举次数，大于n*n退出，好弱啊，水题都a不了//看了别人的代码，我能看出来，这是紫书上给的八数码的方法，同样是看紫书，我为什么领悟不了呢。。。//这个方法思路蛮清晰的诶int n;char a[maxn][maxn];int ans, cnt;bool check(int x, int y) {bool ok = true;if (a[x][y] == 'X' || a[x][y] == '1') return false;int t = x;while (t - 1 >= 0 && a[t - 1][y] != 'X') {t--;if (a[t][y] == '1') return false;}t = y;while (t - 1 >= 0 && a[x][t - 1] != 'X') {t--;if (a[x][t] == '1') return false;}t = x;while (t + 1 < n && a[t + 1][y] != 'X') {t++;if (a[t][y] == '1') return false;}t = y;while (t + 1 < n && a[x][t + 1] != 'X') {t++;if (a[x][t] == '1') return false;}return true; }void dfs() {ans = max(ans, cnt);//访问时更新最值//printf("ans = %d\n", ans);FOR(i, 0, n) FOR(j, 0, n) {if (check(i, j)) {a[i][j] = '1', cnt++;dfs();//从这进入一个分支a[i][j] = '.', cnt--;}}}int main() {//IN(); OUT();while (SF(n) && n) {ans = cnt = 0;//记录次数与最值FOR(i, 0, n) scanf("%s", a[i]);dfs(); PF(ans);}return 0;}//写了个二进制法，，自己举了十几个样例都能过，，还是WA诶，，伤感。我要去参考下别人的代码了额int n;char a[maxn][maxn];vector<int> ans;int temp[4][4];bool phase(int x, int y) {if (a[x][y] == 'X') return false;int t = x;while (t - 1 >= 0 && a[t - 1][y] != 'X') {t--;if (temp[t][y] == 1) return false;}t = y;while (t - 1 >= 0 && a[x][t - 1] != 'X') {t--;if (temp[x][t] == 1) return false;}t = x;while (t + 1 < n && a[t + 1][y] != 'X') {t++;if (temp[t][y] == 1) return false;}t = y;while (t + 1 < n && a[x][t + 1] != 'X') {t++;if (temp[x][t + 1] == 1) return false;}return true; }bool judge() {FOR(i, 0, n) FOR(j, 0, n) {if (temp[i][j] == 1) {if (!phase(i, j)) return false;}}return true;}int check(int x) {int ret = 0;FOR(i, 0, n) FOR(j, 0, n) {if (x & (1<<(i*n + j))) {temp[i][j] = 1, ret++;//printf("ret = %d\n", ret);}else temp[i][j] = 0;if (i && j) {if (temp[i][j]) {if (temp[i][j - 1] || temp[i - 1][j]) return 0;}}}if (judge()) return ret;else return 0;}int main() {//IN(); OUT();while (SF(n) && n) {ans.clear(); CLEAR(temp, 0);FOR(i, 0, n) scanf("%s", a[i]);ans.push_back(0);for (int i = 0; i < (1 << (n*n)); ++i) {ans.push_back(check(i));}sort(ans.begin(), ans.end());cout << ans.back() << endl;}return 0;}//错误想法int n;int d[][2] = { {1, 0}, {0, 1} };int vis[maxn][maxn];char a[maxn][maxn];bool check(int x, int y) {bool ok = true;if (x < 0 || x >= n || y < 0 || y >= n || a[x][y] == 'X' || vis[x][y] == 1) return false;}bool ok(int x, int y) {int t = x;while (t - 1 >= 0 && a[t - 1][y] != 'X') {t--;if (a[t][y] == '1') return false;}t = y;while (t - 1 >= 0 && a[x][t - 1] != 'X') {t--;if (a[x][t] == '1') return false;}t = x;while (t + 1 < n && a[t + 1][y] != 'X') {t++;if (a[t][y] == '1') return false;}t = y;while (t + 1 < n && a[x][t + 1] != 'X') {t++;if (a[x][t + 1] == '1') return false;}return true; }int dfs(int x, int y, int c) {int cnt = 0, r1 = 1, r2 = 0;FOR(i, 0, 2) {if (check(x + d[i][0], y + d[i][1])) {c + 1;vis[x][y] = 1;r1 += dfs(x + d[i][0], y + d[i][1], c);vis[x][y] = 0;r2 += dfs(x + d[i][0], y + d[i][1], c);}}cnt = max(r1, r2);return cnt;}int main() {IN(); OUT();while (SF(n) && n) {CLEAR(vis, 0);FOR(i, 0, n) scanf("%s", a[i]);PF(dfs(0, 0, 0));}return 0;}//不知道该以什么样的方式来搜索，每走一步再往上下左右？