CF 192 Div2

A. Cakeminator

Brute force, from row and column traversal without strawberry coverage

 

char map[30][30]; int vis[30][30]; int hang[30],lie[30]; int main() {      int n,m,i,j;     cin >> n >> m;     for(i=1; i<=n; i++) {         for(j=1; j<=m; j++) {             cin >> map[i][j];             if(map[i][j] == 'S') {                 hang[i] = 1;                 lie[j] = 1;             }         }     }     for(i=1; i<=n; i++) {         if(hang[i] == 0) {             for(j=1; j<=m; j++) {                 vis[i][j] = 1;             }         }     }     for(i=1; i<=m; i++) {         if(lie[i] == 0) {             for(j=1; j<=n; j++) {                 vis[j][i] = 1;             }         }     }     int cnt = 0;     for(i=1; i<=n; i++) {         for(j=1; j<=m; j++) {             if(vis[i][j] == 1)                 cnt++;         }     }     cout << cnt << endl;     return 0; } char map[30][30];int vis[30][30];int hang[30],lie[30];int main() {    int n,m,i,j;    cin >> n >> m;    for(i=1; i<=n; i++) {        for(j=1; j<=m; j++) {            cin >> map[i][j];            if(map[i][j] == 'S') {                hang[i] = 1;                lie[j] = 1;            }        }    }    for(i=1; i<=n; i++) {        if(hang[i] == 0) {            for(j=1; j<=m; j++) {                vis[i][j] = 1;            }        }    }    for(i=1; i<=m; i++) {        if(lie[i] == 0) {            for(j=1; j<=n; j++) {                vis[j][i] = 1;            }        }    }    int cnt = 0;    for(i=1; i<=n; i++) {        for(j=1; j<=m; j++) {            if(vis[i][j] == 1)                cnt++;        }    }    cout << cnt << endl;    return 0;}

B. Road Construction

The distance from a point to another point is less than or equal to 2. Only one point is centered, and other points are connected only to the graph.

M <n/2 all must have at least one point to be the center point.

 

int vis[1005]; int main() {      int n,m,i,j,a,b;     cin >> n >> m;     for(i=1; i<=m; i++) {         cin >> a >> b;         vis[a] = 1;         vis[b] = 1;     }     int t;     for(i=1; i<=n; i++) {         if(vis[i] == 0){           t = i;           break;         }     }     cout << n-1 << endl;     for(i=1; i<=n; i++) {         if(i != t) cout << i << ' ' << t << endl;     }     return 0; } int vis[1005];int main() {    int n,m,i,j,a,b;    cin >> n >> m;    for(i=1; i<=m; i++) {        cin >> a >> b;        vis[a] = 1;        vis[b] = 1;    }    int t;    for(i=1; i<=n; i++) {        if(vis[i] == 0){          t = i;          break;        }    }    cout << n-1 << endl;    for(i=1; i<=n; i++) {        if(i != t) cout << i << ' ' << t << endl;    }    return 0;} 

C. Purification

It took a long time to start thinking too complicated. Check whether all rows or all columns can be overwritten by '.'. If one row is satisfied, the output can be directly

 

char map[105][105]; int x[105],y[105]; int main() {      int n,i,j;     cin >> n;     for(i=0; i<n; i++) {         cin >> map[i];     }     for(i=0; i<n; i++) {         for(j=0; j<n; j++) {             if(map[i][j] == '.') {                 x[i] = 1;                 y[j] = 1;             }         }     }     int cntx = 0,cnty = 0;     for(i=0; i<n; i++) {         if(x[i] != 0) {             cntx++;         }         if(y[i] != 0) {             cnty++;         }     }     if(cnty <n && cntx <n) {         cout << -1 << endl;         return 0;     }     int flag = 0;     if(cntx == n) {         flag = 1;         for(i=0; i<n; i++) {             for(j=0; j<n; j++) {                 if(map[i][j] == '.') {                     cout << i+1 << ' ' << j+1 << endl;                     break;                 }             }         }     }     if(flag == 1)         return 0;     if(cnty == n) {         for(i=0; i<n; i++) {             for(j=0; j<n; j++) {                 if(map[j][i] == '.') {                     cout << j+1 << ' ' << i+1 << endl;                     break;                 }             }         }     }     return 0; } char map[105][105];int x[105],y[105];int main() {    int n,i,j;    cin >> n;    for(i=0; i<n; i++) {        cin >> map[i];    }    for(i=0; i<n; i++) {        for(j=0; j<n; j++) {            if(map[i][j] == '.') {                x[i] = 1;                y[j] = 1;            }        }    }    int cntx = 0,cnty = 0;    for(i=0; i<n; i++) {        if(x[i] != 0) {            cntx++;        }        if(y[i] != 0) {            cnty++;        }    }    if(cnty <n && cntx <n) {        cout << -1 << endl;        return 0;    }    int flag = 0;    if(cntx == n) {        flag = 1;        for(i=0; i<n; i++) {            for(j=0; j<n; j++) {                if(map[i][j] == '.') {                    cout << i+1 << ' ' << j+1 << endl;                    break;                }            }        }    }    if(flag == 1)        return 0;    if(cnty == n) {        for(i=0; i<n; i++) {            for(j=0; j<n; j++) {                if(map[j][i] == '.') {                    cout << j+1 << ' ' << i+1 << endl;                    break;                }            }        }    }    return 0;}

I have only passed three questions, but I have not read D and E.

After the game, it is said that D is a very watery bfs, so I did not write it.