BFS(最短路) HDU 2612 Find a way

题目传送门

  1 /*
  2     BFS：和UVA_11624差不多，本题就是分别求两个点到KFC的最短路，然后相加求最小值
  3 */
  4 /************************************************
  5 Author        :Running_Time
  6 Created Time  :2015-8-4 19:36:36
  7 File Name     :HDOJ_2612.cpp
  8 ************************************************/
  9
 10 #include <cstdio>
 11 #include <algorithm>
 12 #include <iostream>
 13 #include <sstream>
 14 #include <cstring>
 15 #include <cmath>
 16 #include <string>
 17 #include <vector>
 18 #include <queue>
 19 #include <deque>
 20 #include <stack>
 21 #include <list>
 22 #include <map>
 23 #include <set>
 24 #include <bitset>
 25 #include <cstdlib>
 26 #include <ctime>
 27 using namespace std;
 28
 29 #define lson l, mid, rt << 1
 30 #define rson mid + 1, r, rt << 1 | 1
 31 typedef long long ll;
 32 const int MAXN = 2e2 + 10;
 33 const int INF = 0x3f3f3f3f;
 34 const int MOD = 1e9 + 7;
 35 char maze[MAXN][MAXN];
 36 bool vis[MAXN][MAXN];
 37 int step[MAXN][MAXN];
 38 int step2[MAXN][MAXN];
 39 int dx[4] = {-1, 1, 0, 0};
 40 int dy[4] = {0, 0, -1, 1};
 41 int n, m;
 42 int sx, sy, ex, ey;
 43
 44 bool judge_y(int x, int y)  {
 45     if (x < 1 || x > n || y < 1 || y > m || maze[x][y] == ‘#‘ || vis[x][y]) return false;
 46     return true;
 47 }
 48
 49 bool judge_m(int x, int y)  {
 50     if (x < 1 || x > n || y < 1 || y > m || maze[x][y] == ‘#‘ || vis[x][y])  return false;
 51     return true;
 52 }
 53
 54 void BFS_Y(void)    {
 55     memset (step, 0, sizeof (step));
 56     memset (vis, false, sizeof (vis));
 57     queue<pair<int, int> > Q;   Q.push (make_pair (sx, sy));
 58     vis[sx][sy] = true; step[sx][sy] = 0;
 59     while (!Q.empty ()) {
 60         int x = Q.front ().first, y = Q.front ().second;    Q.pop ();
 61         for (int i=0; i<4; ++i) {
 62             int tx = x + dx[i], ty = y + dy[i];
 63             if (!judge_y (tx, ty))   continue;
 64             Q.push (make_pair (tx, ty));    vis[tx][ty] = true;
 65             step[tx][ty] = step[x][y] + 1;
 66         }
 67     }
 68 }
 69
 70 void BFS_M(void)    {
 71     memset (vis, false, sizeof (vis));
 72     memset (step2, 0, sizeof (step2));
 73     queue<pair<int, int> > Q;   Q.push (make_pair (ex, ey));
 74     vis[ex][ey] = true; step2[ex][ey] = 0;
 75     while (!Q.empty ()) {
 76         int x = Q.front ().first, y = Q.front ().second;    Q.pop ();
 77         for (int i=0; i<4; ++i) {
 78             int tx = x + dx[i], ty = y + dy[i];
 79             if (!judge_m (tx, ty))  continue;
 80             step2[tx][ty] = step2[x][y] + 1;
 81             Q.push (make_pair (tx, ty));    vis[tx][ty] = true;
 82         }
 83     }
 84 }
 85
 86 int main(void)    {     //HDU 2612 Find a way
 87     while (scanf ("%d%d", &n, &m) == 2) {
 88         for (int i=1; i<=n; ++i)    {
 89             scanf ("%s", maze[i] + 1);
 90             for (int j=1; j<=m; ++j)    {
 91                 if (maze[i][j] == ‘Y‘)  sx = i, sy = j;
 92                 if (maze[i][j] == ‘M‘)  ex = i, ey = j;
 93             }
 94         }
 95         BFS_Y ();   BFS_M ();
 96         int ans = INF;
 97         for (int i=1; i<=n; ++i)    {
 98             for (int j=1; j<=m; ++j)    {
 99                 if (maze[i][j] == ‘@‘ && vis[i][j])  {
100                     ans = min (ans, step[i][j] + step2[i][j]);
101                 }
102             }
103         }
104         if (ans == INF) puts ("-1");
105         else    printf ("%d\n", ans * 11);
106     }
107
108     return 0;
109 }