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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#17975 | #2213. Knight | laihaochen# | WA | 4ms | 2136kb | C++17 | 1.8kb | 2022-01-15 15:41:40 | 2022-05-04 16:33:31 |
Judging History
answer
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1005;
int n, m, r, c, sx, sy, ex, ey;
char s[N][N];
int col[N][N];
int dx[] = {1, 1, -1, -1};
int dy[] = {1, -1, 1, -1};
void dfs1 (int x, int y, int ch) {
col[x][y] = ch;
for (int i = 0; i < 4; i++) {
int nx = x + dx[i] * r, ny = y + dy[i] * c;
if (1 <= nx && nx <= n && 1 <= ny && ny <= m && s[nx][ny] != '@' && !col[nx][ny]) {
dfs1 (nx, ny, 3 - ch);
}
}
for (int i = 0; i < 4; i++) {
int nx = x + dx[i] * c, ny = y + dy[i] * r;
if (1 <= nx && nx <= n && 1 <= ny && ny <= m && s[nx][ny] != '@' && !col[nx][ny]) {
dfs1 (nx, ny, 3 - ch);
}
}
}
int sz;
bool vis[N][N];
bool dfs2 (int x, int y) {
vis[x][y] = 1, sz++;
if (x == ex && ey == y) return 1;
for (int i = 0; i < 4; i++) {
int nx = x + dx[i] * r, ny = y + dy[i] * c;
if (1 <= nx && nx <= n && 1 <= ny && ny <= m && s[nx][ny] != '@' && !vis[nx][ny]) {
if (dfs2 (nx, ny)) return 1;
}
}
for (int i = 0; i < 4; i++) {
int nx = x + dx[i] * c, ny = y + dy[i] * r;
if (1 <= nx && nx <= n && 1 <= ny && ny <= m && s[nx][ny] != '@' && !vis[nx][ny]) {
if (dfs2 (nx, ny)) return 1;
}
}
return 0;
}
int main() {
scanf ("%d%d%d%d", &n, &m, &r, &c);
for (int i = 1; i <= n; i++) {
scanf ("%s", s[i] + 1);
for (int j = 1; j <= m; j++) {
if (s[i][j] == 'A') sx = i, sy = j;
if (s[i][j] == 'B') ex = i, ey = j;
}
}
if (n < 500 && m < 500) {
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (!col[i][j] && s[i][j] != '@') dfs1 (i, j, 1);
}
}
bool flag = dfs2 (sx, sy);
if (sz == 1) {
printf ("Bob\n");
} else if (!flag) {
printf ("Alice\n");
} else {
printf (col[sz][sy] == col[ex][ey] ? "Alice\n" : "Bob\n");
}
}
return 0;
}
詳細信息
Test #1:
score: 0
Wrong Answer
time: 4ms
memory: 2136kb