UESTC 149 解救小Q

简单 有意思的迷宫问题

主要看平时标记的习惯吧。。 一开始就跪了，多亏迪哥给了组数据。。

5 5
....L
.###a
a#...
##.##
...Q.

answer

9

  1 #include<iostream>

  2 #include<cstdio>

  3 #include<cstdlib>

  4 #include<cstring>

  5 #include<string>

  6 #include<queue>

  7 #include<algorithm>

  8 #include<map>

  9 #include<iomanip>

 10 #include<climits>

 11 #include<string.h>

 12 #include<stdlib.h>

 13 #define INF 1e11

 14 #define MAXN 60

 15 using namespace std;

 16 

 17 struct node {

 18     int x1, y1;

 19     int x2, y2;

 20     bool ok;

 21 }a[30];

 22 

 23 struct dir{

 24     int x, y;

 25     int step;

 26 };

 27 

 28 int dx[] = { -1, 1, 0, 0 }, dy[] = { 0, 0, 1, -1 };

 29 int T, n, m;

 30 char mp[MAXN][MAXN];

 31 int vis[MAXN][MAXN];

 32 int sx, sy, fx, fy;

 33 int num;

 34 map<node, node> s;

 35 

 36 bool judge(int x, int y)

 37 {

 38     if ( vis[x][y] || x < 0 || y < 0 || x >= n || y >= m)

 39         return true;

 40     return false;

 41 }

 42 

 43 int bfs()

 44 {

 45     queue<dir> q;

 46     q.push({sx,sy,false});

 47     vis[sx][sy] = true;

 48     while (!q.empty())

 49     {

 50         dir now = q.front();

 51         q.pop();

 52         if (now.x == fx && now.y == fy) {

 53             return now.step;

 54         }

 55         //cout << "x = " << now.x;

 56         //cout << "  y = " << now.y << endl;

 57         dir next;

 58         next.step = now.step + 1;

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

 60             next.x = now.x + dx[i];

 61             next.y = now.y + dy[i];

 62             if (judge(next.x,next.y)) continue;

 63             if (mp[next.x][next.y] == '#')  continue;

 64             if (mp[next.x][next.y] == '.' || mp[next.x][next.y] == 'Q') {

 65                 q.push({ next.x, next.y,next.step});

 66                 vis[next.x][next.y] = true;

 67 

 68             }

 69             if (mp[next.x][next.y] >= 'a' && mp[next.x][next.y] <= 'z') {

 70                 //cout << "ok" << endl;

 71                 int k = mp[next.x][next.y] - 'a';

 72                 if (next.x == a[k].x1 && next.y == a[k].y1) {

 73                     q.push({ a[k].x2, a[k].y2, next.step });

 74                     //vis[a[k].x1][a[k].y1] = true;     错误的标记

 75                     vis[next.x][next.y] = true;

 76                 }

 77                 else {

 78                     q.push({ a[k].x1, a[k].y1, next.step});

 79                     //vis[next.x][next.y] = true;        错误的标记

 80                     vis[a[k].x2][a[k].y2] = true;

 81                 }

 82             }

 83         }

 84     }

 85     return -1;

 86 }

 87 

 88 void init()

 89 {

 90     memset(a, 0, sizeof(a));

 91     memset(vis, 0, sizeof(vis));

 92     num = 0;

 93 }

 94 

 95 void process()

 96 {

 97     init();

 98     cin >> n >> m;

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

100         cin >> mp[i];

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

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

103             if (mp[i][j] == 'Q')

104                 fx = i, fy = j;

105             else if (mp[i][j] == 'L')

106                 sx = i, sy = j;

107             else if (mp[i][j] >= 'a' && mp[i][j] <= 'z'){

108                 int k = mp[i][j] - 'a';

109                 if (a[k].ok == false) {

110                     a[k].x1 = i;

111                     a[k].y1 = j;

112                     a[k].ok = true;

113                 }

114                 else {

115                     a[k].x2 = i;

116                     a[k].y2 = j;

117                 }

118             }

119         }

120     //cout << "sx = " << sx << " sy =" << sy << endl;

121     //cout << "fx = " << fx << " fy =" << fy << endl;

122     cout << bfs() << endl;

123 }

124 

125 int main()

126 {

127     cin >> T;

128     while (T--)

129         process();

130     //system("pause");

131     return 0;

132 }