HDU 2102 A计划 (BFS或DFS)

时间：2023-03-10 02:14:45
HDU 2102 A计划 (BFS或DFS)

题意：中文题。

析：是一个简单的搜索，BFS 和 DFS都可行， 主要是这个题有一个坑点，那就是如果有一层是#，另一个层是#或者*，都是过不去的，就可以直接跳过，

剩下的就是一个简单的搜索，只不过是两层而已，再加一维表示是哪一层就好，可能一个就是在#必须传送，不能再上下左右走，这个题目已经说的很清楚了，不算坑。

对于BFS，如果找到P就可以结束，如果超过了T，那么就是找不到了，返回找不到。

对于DFS，同样是找到就结束。

总体来说不难。。。。

代码如下：

BFS：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 10 + 10;

const int mod = 1000000007;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r >= 0 && r < n && c >= 0 && c < m;

}

char s[2][maxn][maxn];

int vis[2][maxn][maxn];

int t;

struct Node{

  int pos, x, y;

  Node() { }

  Node(int p, int xx, int yy) : pos(p), x(xx), y(yy) { }

  bool operator == (const Node &p) const{

    return pos == p.pos && x == p.x && y == p.y;

  }

};

bool bfs(){

  memset(vis, -1, sizeof vis);

  Node goal;

  for(int i = 0; i < n; ++i)

    for(int j = 0; j < m; ++j)

      if(s[0][i][j] == 'P')  goal = Node(0, i, j);

      else if(s[1][i][j] == 'P')  goal = Node(1, i, j);

  queue<Node> q;

  vis[0][0][0] = 0;

  q.push(Node(0, 0, 0));

  while(!q.empty()){

    Node u = q.front();  q.pop();

    if(vis[u.pos][u.x][u.y] > t)  return false;

    if(u == goal)  return true;

    for(int i = 0; i < 4; ++i){

      int x = u.x + dr[i];

      int y = u.y + dc[i];

      if(!is_in(x, y) || vis[u.pos][x][y] != -1 || s[u.pos][x][y] == '*')  continue;

      vis[u.pos][x][y] = vis[u.pos][u.x][u.y] + 1;

      if(s[u.pos][x][y] == '#'){

        vis[u.pos^1][x][y] = vis[u.pos][x][y];

        if(s[u.pos^1][x][y] == '#' || s[u.pos^1][x][y] == '*')  continue;

        q.push(Node(u.pos^1, x, y));

        continue;

      }

      q.push(Node(u.pos, x, y));

    }

  }

  return false;

}

int main(){

  int T;  cin >> T;

  while(T--){

    scanf("%d %d %d", &n, &m, &t);

    for(int i = 0; i < n; ++i)  scanf("%s", s[0]+i);

    for(int i = 0; i < n; ++i)  scanf("%s", s[1]+i);

    bool ans = bfs();

    printf("%s\n", ans ? "YES" : "NO");

  }

  return 0;

}

　DFS：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 10 + 10;

const int mod = 1000000007;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r >= 0 && r < n && c >= 0 && c < m;

}

char s[2][maxn][maxn];

bool vis[2][maxn][maxn];

int t;

bool dfs(int pos, int r, int c, int d){

  if(d > t)  return false;

  if(s[pos][r][c] == 'P')  return true;

  for(int i = 0; i < 4; ++i){

    int x = r + dr[i];

    int y = c + dc[i];

    if(!is_in(x, y) || vis[pos][x][y] || s[pos][x][y] == '*')  continue;

    vis[pos][x][y] = 1;

    if(s[pos][x][y] == '#'){

      vis[pos^1][x][y] = 1;

      if(dfs(pos^1, x, y, d+1))  return true;

      vis[pos^1][x][y] = 0;

    }

    else if(dfs(pos, x, y, d+1))  return true;

    vis[pos][x][y] = 0;

  }

  return false;

}

int main(){

  int T;  cin >> T;

  while(T--){

    scanf("%d %d %d", &n, &m, &t);

    for(int i = 0; i < n; ++i)  scanf("%s", s[0]+i);

    for(int i = 0; i < n; ++i)  scanf("%s", s[1]+i);

    for(int i = 0; i < n; ++i)

      for(int j = 0; j < m; ++j)

        if(s[0][i][j] == '#' && (s[1][i][j] == '*' || s[1][i][j] == '#'))  s[0][i][j] = s[1][i][j] = '*';

        else if(s[0][i][j] == '*' && s[1][i][j] == '#')  s[1][i][j] = '*';

    memset(vis, 0, sizeof vis);

    vis[0][0][0] = 1;

    bool ans = dfs(0, 0, 0, 0);

    printf("%s\n", ans ? "YES" : "NO");

  }

  return 0;

}

　　

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