Codeforces 106D Treasure Island preprocessing prefix and + brute force (Water

Codeforces 106D Treasure Island preprocessing prefix and + brute force (Water

Question link: Click the open link

Question:

Matrix given n * m

# The walls and letters are flat.

A maximum of 26 letters are allowed (not repeated)

The following k commands,

Each instruction represents the movement direction and number of steps.

If you start with a letter and execute all the commands in sequence, and no process will hit the wall or go out of the map, the letter is valid.

Outputs all valid letters in Lexicographic Order. If no letter is valid, 'no solution' is output'

Prefix and then brute force.

#include 
 
  #include 
  
   #include #include 
   
    #include 
    
     #include 
     
      #include 
      using namespace std;#define N 1005vector
       
        ans;struct node{ int x, y; char c; void put(){printf("(%d,%d) : %c\n", x, y, c);}}a[30];int step[4][2] = {-1,0, 1,0, 0,-1, 0,1};int work[100005][2];char s[N];int mp[N][N], n, m, k, top, h[N][N], l[N][N];bool okh(int H, int x, int y){return h[H][y]-h[H][x-1] == 0;}bool okl(int L, int x, int y){return l[L][y]-l[L][x-1] == 0;}bool ok(int x, int y, int i, int j){ if(x>i)swap(x,i); if(y>j)swap(y,j); if(x==i) return okh(x, y, j); else return okl(y, x, i);}bool inmap(int x, int y){return 1<=x&&x<=n&&1<=y&&y<=m;}bool judge(int x, int y){ for(int i = 0; i < k; i++) { int ux = step[work[i][0]][0] * work[i][1] + x, uy = step[work[i][0]][1] *work[i][1] + y; if(!inmap(ux,uy))return false; if(!ok(x, y, ux, uy))return false; x = ux; y = uy; } return true;}void debug(){for(int i = 0; i < top; i++)a[i].put();}void solve(){ // debug(); ans.clear(); for(int i = 0; i < top; i++) if(judge(a[i].x, a[i].y)) ans.push_back(a[i].c); if(ans.size()==0){ puts("no solution"); return ; } sort(ans.begin(), ans.end()); for(int i = 0; i < ans.size(); i++)printf("%c",ans[i]); puts("");}void input(){ top = 0; memset(mp, 0, sizeof mp); memset(h, 0, sizeof h); memset(l, 0, sizeof l); for(int i = 1; i <= n; i++){ scanf("%s", s+1); for(int j = 1; j <= m; j++) { if(s[j]=='#') { mp[i][j] = 1; h[i][j] = 1; l[j][i] = 1; } else if('A'<=s[j] && s[j]<='Z'){ a[top].x = i; a[top].y = j; a[top].c = s[j]; top++; } } } for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) h[i][j] += h[i][j-1]; for(int i = 1; i <= m; i++) for(int j = 1; j <= n; j++) l[i][j] += l[i][j-1]; scanf("%d",&k); for(int i = 0; i < k; i++){ scanf("%s %d", s, &work[i][1]); if(s[0] == 'N') work[i][0] = 0; else if(s[0]=='S') work[i][0] = 1; else if(s[0]=='W') work[i][0] = 2; else work[i][0] = 3; }}int main(){ while(~scanf("%d %d",&n,&m)){ input(); solve(); } return 0;}