hihoCoder #1092 Have Launch Together

时间：2015-04-07 13:45:21 阅读：178 评论：0 收藏：0 [点我收藏+]

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Description

Everyday Littile Hi and Little Ho meet in the school cafeteria to have lunch together. The cafeteria is often so crowded that two adjacent seats are hard to find.

School cafeteria can be considered as a matrix of N*M blocks. Each block can be empty or occupied by people, obstructions and seats. Little Hi and Little Ho starts from the same block. They need to find two adjacent seats(two seats are adjacent if and only if their blocks share a common edge) without passing through occupied blocks. Further more, they want the total distance to the seats is minimal.

Little Hi and Little Ho can move in 4 directions (up, down, left, right) and they can not move outside the matrix.

Input

Input cantains a single testcase.

First line contains two integers N, M, the length and width of school cafeteria.

The next is a matrix consists of N lines, each line containing M characters. Each character describe a block: ‘.‘ for empty block, ‘P‘ for people, ‘#‘ for obstructions, ‘S‘ for seats and ‘H‘ for Little Hi and Little Ho‘s starting position.

10 <= N, M <= 100

Output

Output the minimal distance they need to move to reach two adjacent seats. If no such adjacent seats output a line "Hi and Ho will not have lunch." without quotes.

Sample Input

10 10
##########
#...P##..#
#S#...#.P#
#S#..#...#
#...#.####
#.#...#.H#
##......##
#..P#..S.#
##.......#
##########

Sample Output

Solution:

最短路径算法的简单应用。TALK IS CHEAP......

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <climits>
  5 #include <cmath>
  6 using namespace std;
  7 
  8 char g[100][100];
  9 int N, M;
 10 
 11 int d[100][100];
 12 bool spt[100][100];
 13 int Hi,Hj;
 14 
 15 int minDistance()
 16 {
 17     int minm = INT_MAX, min_index;
 18     for (int i = 0; i < N; ++i) {
 19         for (int j = 0; j < M; ++j) {
 20             if(spt[i][j] == false && d[i][j] <= minm) 
 21                 minm = d[i][j], min_index = i*N + j;
 22         }
 23     }
 24 
 25     return min_index;
 26 }
 27 
 28 void dijkstra()
 29 {
 30     for (int i = 0; i < N; ++i) 
 31         for (int j = 0; j < M; ++j) {
 32             d[i][j] = INT_MAX, spt[i][j] = false;
 33         }
 34 
 35     d[Hi][Hj] = 0;
 36     int V = N*M;
 37     for (int k = 0; k < V-1; ++k) {
 38         int u = minDistance();
 39 
 40         spt[u/N][u%N] = true;
 41 
 42         for (int i = 0; i < N; ++i) 
 43             for (int j = 0; j < M; ++j) {
 44                 if (!spt[i][j] && ((g[i][j] == ‘.‘) && abs(i-u/N) + abs(j-u%N) == 1)
 45                     && d[u/N][u%N] != INT_MAX && (d[u/N][u%N] + 1 < d[i][j]) ) {
 46                     d[i][j] = d[u/N][u%N] + 1;
 47                 }
 48             }
 49     }
 50 }
 51 
 52 
 53 bool hasSeats()
 54 {
 55     for (int i = 0; i < N; ++i) {
 56         for (int j = 0; j < M; ++j) {
 57             if (g[i][j] == ‘S‘) {
 58                 if ((i+1 < N && g[i+1][j] == ‘S‘)
 59                     || (j+1 < M && g[i][j+1] == ‘S‘)
 60                     || (i-1 >= 0 && g[i-1][j] == ‘S‘)
 61                     || (j-1 >= 0 && g[i][j-1] == ‘S‘)) {
 62                     return true;
 63                 }
 64             }
 65         }
 66     }
 67 
 68     return false;
 69 }
 70 
 71 int solve()
 72 {
 73     int minm = INT_MAX;
 74     for (int i = 0; i < N; ++i) {
 75         for (int j = 0; j < M; ++j) {
 76             if (g[i][j] == ‘S‘) {
 77                 if ((i+1 < N && g[i+1][j] == ‘S‘)) {
 78                     int d1 = INT_MAX;
 79                     if (i-1>=0) {
 80                         d1 = min(d1, d[i-1][j]);
 81                     }
 82                     if (j-1>=0) {
 83                         d1 = min(d1, d[i][j-1]);
 84                     }
 85                     if (j+1<M) {
 86                         d1 = min(d1, d[i][j+1]);
 87                     }
 88 
 89                     int d2 = INT_MAX;
 90                     if (j-1>=0) {
 91                         d2 = min(d2, d[i+1][j-1]);
 92                     }
 93                     if (j+1 < M) {
 94                         d2 = min(d2, d[i+1][j+1]);
 95                     }
 96                     if (i+2 < N) {
 97                         d2 = min(d2, d[i+2][j]);
 98                     }
 99                     if (d1 != INT_MAX && d2 != INT_MAX)
100                         minm = min(d1+d2+2, minm);
101                 }
102                 if (j+1 < M && g[i][j+1] == ‘S‘) {
103                     int d1 = INT_MAX;
104                     int d2 = INT_MAX;
105 
106                     if (i-1>=0) {
107                         d1 = min(d1, d[i-1][j]);
108                     }
109                     if (j-1>=0) {
110                         d1 = min(d1, d[i][j-1]);
111                     }
112                     if (i+1 < N) {
113                         d1 = min(d1, d[i+1][j]);
114                     }
115 
116                     if (i-1>=0) {
117                         d2 = min(d2, d[i-1][j+1]);
118                     }
119                     if (j+2 < M) {
120                         d2 = min(d2, d[i][j+2]);
121                     }
122                     if (i+1 < N) {
123                         d2 = min(d2, d[i+1][j+1]);
124                     }
125 
126                     if (d1 != INT_MAX && d2 != INT_MAX)
127                         minm = min(d1+d2+2, minm);
128 
129                 }
130             }
131         }
132     }
133 
134     return minm;
135 }
136 
137 int main()
138 {
139     scanf("%d %d", &N, &M);
140     for (int i = 0; i < N; ++i) {
141         for (int j = 0; j < M; ++j) {
142             scanf(" %c", g[i] + j);
143             if (g[i][j] == ‘H‘) {
144                 Hi = i, Hj = j;
145             }
146         }
147     }
148 
149     if (!hasSeats()) {
150         cout << "Hi and Ho will not have lunch." << endl;
151     }
152     else {
153         dijkstra();
154 
155         int dist = solve();
156         if (dist == INT_MAX) {
157             cout << "Hi and Ho will not have lunch." << endl;
158         }
159         else {
160             cout << dist << endl;
161         }
162     }
163 }

hihoCoder #1092 Have Launch Together

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原文地址：http://www.cnblogs.com/liew/p/4398015.html

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