AOJ0558 Cheese

問題文
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0558

N個の工場は1種類のみ生産するので,次に行くべき場所が一意に定まる.N回幅優先で最短時間をもとめも十分間に合う.

Code

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#include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <sstream>
#include <map>
#include <set>
#include <queue>

#define REP(i,k,n) for(int i=k;i<n;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1<<30
#define pb push_back
#define mp make_pair

using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int h, w, n;
int dx[4] = {1,0,-1,0};
int dy[4] = {0,1,0,-1};

bool can(int y,int x) {
  if(0 <= y && y < h && 0 <= x && x < w) return true;
  return false;
}

int main() {
  cin >> h >> w >> n;

  vector<string> s(h);
  rep(i, h) cin >> s[i];

  int sx = -1, sy = -1;
  rep(i, h) {
      rep(j, w) {
          if(s[i][j] == 'S') {
              sy = i;
              sx = j;
          }
      }
  }

  int ans = 0, y = sy, x = sx;
  int d[1005][1005];
  bool used[1005][1005];

  rep(i, n) {
      queue<P> que;
      que.push(P(y, x));

      memset(used, 0, sizeof(used));
      used[y][x] = x;

      memset(d, 0, sizeof(d));
      int cnt = 0;

      while(que.size()) {
          P p = que.front();
          que.pop();

          if(s[p.first][p.second] == char('1' + i)) {
              ans += d[p.first][p.second];
              y = p.first;
              x = p.second;
              break;
          }

          rep(j, 4) {
              int ny = p.first + dy[j];
              int nx = p.second + dx[j];

              if(can(ny, nx) && !used[ny][nx] && s[ny][nx] != 'X') {
                  used[ny][nx] = true;
                  d[ny][nx] += d[p.first][p.second] + 1;
                  que.push(P(ny, nx));
              }
          }
      }
  }

  cout << ans << endl;
  return 0;
}
Feb 12th, 2016