c++ - 来自多个来源的 BFS 上的 TLE

Question

我正在尝试解决这个问题:http://olimpiada-informatica.org/?cmd=downloadE&pbm=velo101&ext=pdf它是西类牙语，但我会尝试在这里翻译它:

You are about to land on earth when you find out that someone has released some Velociraptors in the landing strip.
The landing trip has a rectangular shape. We mark with a dot (.)an empty spot, with a V the velociraptors and with a # the spots with obstacles on them (you cannot land on them).
It takes a second to the velociraptors to go to another spot, but they can only move horizontally and vertically.
You are asked to mark with an X those spots in which, landing on them, you would maximize your remaining lifetime.

我已经完成了一个算法，在该算法中我占据了迅猛龙的每个位置并在每个位置上进行了 BFS，但是我得到了 TLE，这是我的代码:

http://ideone.com/a6BVv3

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <utility>
using namespace std;

int cost[501][501],xx,yy,n,m;
char mat[501][501];
bool visit[501][501],first = true;

int a[] = {-1,0,0,1}, b[] = {0,-1,1,0};

void check(int x,int y,int level) {
  cost[x][y] = level;
  for(int i = 0; i < 4; ++i) {
    xx = x + a[i];
    yy = y + b[i];
    if(0 <= xx and xx < n and 0 <= yy and yy < m and mat[xx][yy] == '.') {
      if(!visit[xx][yy] or level + 1 < cost[xx][yy]) {
        visit[xx][yy] = true;
        check(xx,yy,level + 1);
      }
    }
  }
}

int max() {
  int r = -1;
  for(int i = 0; i < n; ++i) for(int j = 0; j < m; ++j) if(mat[i][j] == '.') r = max(r,cost[i][j]);
  return r;
}

void show() {
  if(!first) puts("---");
  int r = max();
  for(int i = 0; i < n; ++i) {
    for(int j = 0; j < m; ++j) {
      if(cost[i][j] == r) printf("X");
      else printf("%c",mat[i][j]);
    }
    puts("");
  }
}

int main() {
  while(scanf("%d %d",&n,&m) == 2) {
    queue<pair<int,int> > cola;
    for(int i = 0; i < n; ++i) {
      scanf("\n");
      for(int j = 0; j < m; ++j) {
        scanf("%c",&mat[i][j]);
        if(mat[i][j] == 'V') cola.push(make_pair(i,j));
      }
    }
    memset(cost,-1,sizeof cost);
    memset(visit,0,sizeof visit);
    while(!cola.empty()) {
      pair<int,int> aux = cola.front();
      visit[aux.first][aux.second] = true;
      check(aux.first, aux.second,0);
      cola.pop();
    }
    show();
    first = false;
  }
  return 0;
}

有人知道如何改进我的算法吗？

编辑

好的，我能够解决这个问题，如果有人感兴趣的话，这里是代码:

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <utility>
using namespace std;

int cost[501][501],n,m;
char mat[501][501];
bool visit[501][501],first = true;
queue<pair<int,int> > cola;

int a[] = {-1,0,0,1}, b[] = {0,-1,1,0};

int max() {
  int r = -1;
  for(int i = 0; i < n; ++i) for(int j = 0; j < m; ++j) if(mat[i][j] == '.') r = max(r,cost[i][j]);
  return r;
}

void show() {
  if(!first) puts("---");
  int r = max();
  for(int i = 0; i < n; ++i) {
    for(int j = 0; j < m; ++j) {
      if(cost[i][j] == r) printf("X");
      else printf("%c",mat[i][j]);
    }
    puts("");
  }
}

int main() {
  int cont = 0,x,y,xx,yy,level;
  while(scanf("%d %d",&n,&m) == 2) {
    for(int i = 0; i < n; ++i) {
      scanf("\n");
      for(int j = 0; j < m; ++j) {
        scanf("%c",&mat[i][j]);
        if(mat[i][j] == 'V') cola.push(make_pair(i,j));
      }
    }
    memset(cost,-1,sizeof cost);
    memset(visit,0,sizeof visit);
    while(!cola.empty()) {
      int s_cola = cola.size();
      for(int i = 0; i < s_cola; ++i) {
        x = cola.front().first, y = cola.front().second;
        cola.pop();
        level = cost[x][y];
        for(int i = 0; i < 4; ++i) {
          xx = x + a[i], yy = y + b[i];
          if(0 <= xx and xx < n and 0 <= yy and yy < m and mat[xx][yy] == '.') {
            if(!visit[xx][yy] or level + 1 < cost[xx][yy]) {
              visit[xx][yy] = true;
              cost[xx][yy] = level + 1;
              cola.push(make_pair(xx,yy));
            }
          }
        }
      }
    }
    show();
    first = false;
  }
  return 0;
}

Answer 1

您正在 check() 中对整个图进行深度优先搜索。将其与 main() 中的循环集成，而不是尝试以深度优先的方式寻找最短路径。