深搜简单题

城堡问题

题目传送：2815:城堡问题

AC代码（递归形式的dfs）：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n, m;
int room[55][55];
int color[55][55];
int color_num, max_room;
int tmp_room;

void dfs(int i, int j) {
    if(color[i][j]) {
        return;
    }
    color[i][j] = color_num;
    tmp_room ++;
    if((room[i][j] & 1) == 0) dfs(i, j - 1);//巧妙运用位运算，来判断是否有墙，且这里位运算要加上括号，因为位运算优先级较低 
    if((room[i][j] & 2) == 0) dfs(i - 1, j);
    if((room[i][j] & 4) == 0) dfs(i, j + 1);
    if((room[i][j] & 8) == 0) dfs(i + 1, j);
}

int main() {
    while(scanf("%d %d", &n, &m) != EOF) {
        for(int i = 0; i < n; i ++) {
            for(int j = 0; j < m; j ++) {
                scanf("%d", &room[i][j]);
            }
        }
        color_num = 0, max_room = 0;
        memset(color, 0, sizeof(color));
        for(int i = 0; i < n; i ++) {
            for(int j = 0; j < m; j ++) {
                tmp_room = 0;//判断每一个连通分量（即房间）里的大小 
                if(color[i][j] == 0) {
                    color_num ++;
                    dfs(i, j);
                    max_room = max(max_room, tmp_room);//取最大值 
                }
            }
        }
        printf("%d\n%d\n", color_num, max_room);
    }
    return 0;
}

AC代码（非递归形式的dfs）：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n, m;
int room[55][55];
int color[55][55];
int color_num, max_room;
int tmp_room;

struct Room {
    int i, j;
    Room(int _i, int _j) : i(_i), j(_j) {
    } 
};

void dfs(int i, int j) {//用stack模拟的非递归dfs 

    stack stk;
    stk.push(Room(i,j));

    while(!stk.empty()) {
        Room rm = stk.top();
        int p = rm.i, q = rm.j;
        if(color[p][q]) stk.pop();
        else {
            tmp_room ++;
            color[p][q] = color_num;
            if((room[p][q] & 1) == 0) stk.push(Room(p, q - 1));
            if((room[p][q] & 2) == 0) stk.push(Room(p - 1, q));
            if((room[p][q] & 4) == 0) stk.push(Room(p, q + 1));
            if((room[p][q] & 8) == 0) stk.push(Room(p + 1, q));
        }
    }
}

int main() {
    while(scanf("%d %d", &n, &m) != EOF) {
        for(int i = 0; i < n; i ++) {
            for(int j = 0; j < m; j ++) {
                scanf("%d", &room[i][j]);
            }
        }
        color_num = 0, max_room = 0;
        memset(color, 0, sizeof(color));
        for(int i = 0; i < n; i ++) {
            for(int j = 0; j < m; j ++) {
                tmp_room = 0;//判断每一个连通分量（即房间）里的大小 
                if(color[i][j] == 0) {
                    color_num ++;
                    dfs(i, j);
                    max_room = max(max_room, tmp_room);//取最大值 
                }
            }
        }
        printf("%d\n%d\n", color_num, max_room);
    }
    return 0;
}

---

Sudoku

题目传送：POj - 2676 - Sudoku

AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define INF 0x7fffffff
using namespace std;

int mp[10][10];//代表整个图 
int row_flag[10][10];//表示第i行存不存在j这个数字 
int col_flag[10][10];//表示第i列存不存在j这个数字 
int block_flag[10][10];//表示第i宫格存不存在j这个数字

struct Pos {//图上一个位置 
    int x, y;
    Pos(int _x, int _y) : x(_x), y(_y) {
    } 
};

int get_block(int i, int j) {//获得一个位置所在的宫格 
    int x = (i - 1) / 3;
    int y = (j - 1) / 3;
    return x * 3 + y + 1;
}

void set_flag(int i, int j, int num, int f) {//为每一个位置设置标志 
    row_flag[i][num] = f;
    col_flag[j][num] = f;
    block_flag[get_block(i, j)][num] = f;
}

bool is_ok(int i, int j, int num) {
    if(!row_flag[i][num] && !col_flag[j][num] && !block_flag[get_block(i, j)][num]) return true;
    return false;
}

vector blank_pos;

bool dfs(int n) {
    if(n < 0) return true;//递归边界 

    int x = blank_pos[n].x;
    int y = blank_pos[n].y;
    for(int i = 1; i <= 9; i ++) {
        if(is_ok(x, y, i)) {
            mp[x][y] = i;
            set_flag(x, y, i, 1);
            if(dfs(n - 1)) return true;
            set_flag(x, y, i, 0);//回溯 
        }
    }
    return false;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T --) {

        memset(row_flag, 0, sizeof(row_flag));
        memset(col_flag, 0, sizeof(col_flag));
        memset(block_flag, 0, sizeof(block_flag));
        blank_pos.clear();

        for(int i = 1; i <= 9; i ++) {
            char str[15];
            scanf("%s", str + 1);
            for(int j = 1; j <= 9; j ++) {
                mp[i][j] = str[j] - '0';
                if(mp[i][j] != 0) {
                    set_flag(i, j, mp[i][j], 1);
                }
                else {
                    blank_pos.push_back(Pos(i, j));
                }
            }
        }

        if(dfs(blank_pos.size() - 1)) {
            for(int i = 1; i <= 9; i ++) {
                for(int j = 1; j <= 9; j ++) {
                    printf("%d", mp[i][j]);
                }
                printf("\n");
            }
        }
    }
    return 0;
}

---

Find The Multiple

题目传送：POJ - 1426 - Find The Multiple

这个题目注意到因为最多到200，打个表发现longlong可以存下答案，dfs即可

AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n;

bool dfs(LL ans, int ceng) {
    if(ans % n == 0) {
        printf("%I64d\n", ans);
        return true;
    }
    if(ceng == 19) {
        return false;
    }
    if(dfs(ans * 10, ceng + 1)) return true;
    if(dfs(ans * 10 + 1, ceng + 1)) return true;
    return false;
}

int main() {
    while(scanf("%d", &n) != EOF) {
        if(n == 0) break;
        dfs(1LL, 1);
    }
    return 0;
}

---