CodeCraft-20 (Div. 2)

D

被样例的构造误导到了奇怪的地方，以为“无限走”就要不断经过起始格子，实际上可以走到某个“核心”处然后在核心处绕圈圈。

int n;
int x[1005][1005];
int y[1005][1005];

int vis[1005][1005];
int deg[1005][1005];
char ans[1005][1005];

int dx[4] = {0, 0, 1, -1};
int dy[4] = {-1, 1, 0, 0};
char dans[4] = {'R', 'L', 'U', 'D'};
char adans[4] = {'L', 'R', 'D', 'U'};

queue<pii> Q;
void bfs1(int ux, int uy) {
    vis[ux][uy] = 1;
    ans[ux][uy] = 'X';
    Q.push({ux, uy});
    while(!Q.empty()) {
        int vx = Q.front().first;
        int vy = Q.front().second;
        Q.pop();
        for(int i = 0; i < 4; ++i) {
            int tx = vx + dx[i];
            int ty = vy + dy[i];
            if(1 <= tx && tx <= n && 1 <= ty && ty <= n) {
                if(x[tx][ty] == ux && y[tx][ty] == uy && vis[tx][ty] == 0) {
                    vis[tx][ty] = 1;
                    ans[tx][ty] = dans[i];
                    Q.push({tx, ty});
                }
            }
        }
    }
}

void bfs2(int ux, int uy) {
    vis[ux][uy] = 1;
    Q.push({ux, uy});
    for(int i = 0; i < 4; ++i) {
        int tx = ux + dx[i];
        int ty = uy + dy[i];
        if(1 <= tx && tx <= n && 1 <= ty && ty <= n) {
            if(x[tx][ty] == -1 && y[tx][ty] == -1 && vis[tx][ty] == 0) {
                vis[tx][ty] = 1;
                ans[ux][uy] = adans[i];
                ans[tx][ty] = dans[i];
                Q.push({tx, ty});
                break;
            }
        }
    }
    if(Q.size() == 1) {
        puts("INVALID");
        exit(0);
    }
    while(!Q.empty()) {
        int vx = Q.front().first;
        int vy = Q.front().second;
        Q.pop();
        for(int i = 0; i < 4; ++i) {
            int tx = vx + dx[i];
            int ty = vy + dy[i];
            if(1 <= tx && tx <= n && 1 <= ty && ty <= n) {
                if(x[tx][ty] == -1 && y[tx][ty] == -1 && vis[tx][ty] == 0) {
                    vis[tx][ty] = 1;
                    ans[tx][ty] = dans[i];
                    Q.push({tx, ty});
                }
            }
        }
    }
}

void test_case() {
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= n; ++j) {
            scanf("%d%d", &x[i][j], &y[i][j]);
            ans[i][j] = '.';
        }
    }

    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= n; ++j) {
            if(vis[i][j] == 0 && x[i][j] == i && y[i][j] == j)
                bfs1(i, j);
        }
    }

    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= n; ++j) {
            if(vis[i][j] == 0 && x[i][j] == -1 && y[i][j] == -1)
                bfs2(i, j);
        }
    }

    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= n; ++j) {
            if(vis[i][j] == 0) {
                puts("INVALID");
                exit(0);
            }
        }
    }

    puts("VALID");
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= n; ++j)
            putchar(ans[i][j]);
        putchar('\n');
    }
}

来源：https://www.cnblogs.com/KisekiPurin2019/p/12418168.html