ural 1338. Automobiles

1338. Automobiles

Time limit: 3.0 second
Memory limit: 64 MB
Everyone knows what traffic congestion is. Even supercomputer won’t be able to lay an optimal passage of the city center in the rush-hours. But it is possible to simulate the traffic flows.
For that the following experiment is carried out. We choose several check points on the streets net of the city and one goal T from the checkpoints. We send a car from each checkpoint (except T) to the point T along the shortest way. In the point T we fix how many cars came from the North, how many from the South, from the East and from the West. So we can judge about the functioning capacity of the approaches to the point T.
You are to carry out such an experiment. No, you are not given a car! You are no to go anywhere. You are only to write a program simulating the experiment.

Input

Input contains the city plan description in the following format. The first line contains two integers W and H (1 ≤ W, H ≤ 500) – the width and the height of the plan. The next H lines consists of the streets net and checkpoints description. A symbol “.” means a place where a building is. A symbol “#” means a road fragment. A symbol “o” (small Latin letter) means a checkpoint. A road fragment always occupy a cell completely. Two road fragments belong to one road if and only if they have a common side.
Then there is a series of assignments of the experiment. First of all there is a number of assignments M (0 ≤ M ≤ 20). Each of the next M lines contains the number of the goal point T for the corresponding experiment. Assume that the checkpoints are numbered bottom-up and from left to right.
If some car is to choose a way from several shortest ones the next scheme of priorities acts: South, North, West, East.

Output

You are to output the results of each experiment in the following format:
Experiment #N: North: Rn, South: Rs, East: Re, West: Rw
where Rn, Rs, Re and Rw - an amount of cars that came in the experiment number N to the goal point from the North, South, East and West respectively.

Sample

inputoutput
10 5
..####....
..o..o....
..####.#o.
......##..
.o#####...
1
4

Experiment #1: North: 0, South: 1, East: 0, West: 0
Problem Author: Alexander Klepinin
Problem Source: USU Championship 2004
Difficulty: 1752
 
题意：给出一个图o为检查点，可走，.不可走，#可走
首先要求最短，其次有一个对于方向的优先顺序South, North, West, East.
问对于某个检查点，其他检查点按照上述规则走，是从哪个方向走到这个点。
分析：
这题居然是居然难度这么高。。。
不就是bfs吗。。
注意优先顺序
  1 /**
  2 Create By yzx - stupidboy
  3 */
  4 #include <cstdio>
  5 #include <cstring>
  6 #include <cstdlib>
  7 #include <cmath>
  8 #include <deque>
  9 #include <vector>
 10 #include <queue>
 11 #include <iostream>
 12 #include <algorithm>
 13 #include <map>
 14 #include <set>
 15 #include <ctime>
 16 #include <iomanip>
 17 using namespace std;
 18 typedef long long LL;
 19 typedef double DB;
 20 #define For(i, s, t) for(int i = (s); i <= (t); i++)
 21 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
 22 #define Rep(i, t) for(int i = (0); i < (t); i++)
 23 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
 24 #define rep(i, x, t) for(int i = (x); i < (t); i++)
 25 #define MIT (2147483647)
 26 #define INF (1000000001)
 27 #define MLL (1000000000000000001LL)
 28 #define sz(x) ((int) (x).size())
 29 #define clr(x, y) memset(x, y, sizeof(x))
 30 #define puf push_front
 31 #define pub push_back
 32 #define pof pop_front
 33 #define pob pop_back
 34 #define ft first
 35 #define sd second
 36 #define mk make_pair
 37 inline void SetIO(string Name)
 38 {
 39     string Input = Name+".in",
 40     Output = Name+".out";
 41     freopen(Input.c_str(), "r", stdin),
 42     freopen(Output.c_str(), "w", stdout);
 43 }
 44 
 45 
 46 inline int Getint()
 47 {
 48     int Ret = 0;
 49     char Ch = ' ';
 50     bool Flag = 0;
 51     while(!(Ch >= '0' && Ch <= '9'))
 52     {
 53         if(Ch == '-') Flag ^= 1;
 54         Ch = getchar();
 55     }
 56     while(Ch >= '0' && Ch <= '9')
 57     {
 58         Ret = Ret * 10 + Ch - '0';
 59         Ch = getchar();
 60     }
 61     return Flag ? -Ret : Ret;
 62 }
 63 
 64 const int N = 510;
 65 const int Dx[] = {1, -1, 0, 0}, Dy[] = {0, 0, -1, 1};
 66 const int Where[] = {1, 0, 3, 2};
 67 int m, n;
 68 char Map[N][N];
 69 typedef pair<int, int> II;
 70 II Point[N * N], Que[N * N * 4];
 71 int Tot;
 72 int Dp[N][N], Mark[N][N], Come[N][N];
 73 
 74 inline void Input()
 75 {
 76     scanf("%d%d", &m, &n);
 77     For(i, 1, n)
 78     {
 79         getchar();
 80         scanf("%s", Map[i] + 1);
 81     }
 82 }
 83 
 84 inline bool Check(int x, int y)
 85 {
 86     if(x < 1 || x > n || y < 1 || y > m) return 0;
 87     if(Map[x][y] == '.') return 0;
 88     return 1;
 89 }
 90 
 91 inline void Bfs(int Stx, int Sty)
 92 {
 93     For(i, 1, n)
 94         For(j, 1, m)
 95             Dp[i][j] = INF, Mark[i][j] = Come[i][j] = -1;
 96     int Tx, Ty, Head = 1, Tail = 0;
 97     Rep(i, 4)
 98     {
 99         Tx = Stx + Dx[i];
100         Ty = Sty + Dy[i];
101         if(Check(Tx, Ty))
102         {
103             Dp[Tx][Ty] = 1;
104             Mark[Tx][Ty] = i;
105             Come[Tx][Ty] = Where[i];
106             Que[++Tail] = mk(Tx, Ty);
107         }
108     }
109 
110     while(Head <= Tail)
111     {
112         II Now = Que[Head++];
113         int x = Now.ft, y = Now.sd;
114         Rep(i, 4)
115         {
116             Tx = x + Dx[i];
117             Ty = y + Dy[i];
118             if(Check(Tx, Ty))
119             {
120                 if(Dp[Tx][Ty] > Dp[x][y] + 1 ||
121                     (Dp[Tx][Ty] == Dp[x][y] + 1 && Come[Tx][Ty] > Where[i]))
122                 {
123                     Dp[Tx][Ty] = Dp[x][y] + 1;
124                     Mark[Tx][Ty] = Mark[x][y];
125                     Come[Tx][Ty] = Where[i];
126                     Que[++Tail] = mk(Tx, Ty);
127                 }
128             }
129         }
130     }
131 }
132 
133 inline void Solve()
134 {
135     Ford(i, n, 1)
136         For(j, 1, m)
137             if(Map[i][j] == 'o')
138                 Point[++Tot] = mk(i, j);
139 
140     int Opt, Ans[4];
141     scanf("%d", &Opt);
142     For(Test, 1, Opt)
143     {
144         int x;
145         scanf("%d", &x);
146         Bfs(Point[x].ft, Point[x].sd);
147 
148         Rep(i, 4) Ans[i] = 0;
149         For(i, 1, Tot)
150             if(x != i && Mark[Point[i].ft][Point[i].sd] != -1)
151                 Ans[Mark[Point[i].ft][Point[i].sd]]++;
152         printf("Experiment #%d: North: %d, South: %d, East: %d, West: %d\n",
153                 Test, Ans[1], Ans[0], Ans[3], Ans[2]);
154     }
155 }
156 
157 int main()
158 {
159     #ifndef ONLINE_JUDGE
160     SetIO("A");
161     #endif
162     Input();
163     Solve();
164     return 0;
165 }
View Code

 

posted @ 2015-11-26 21:55  yanzx6  阅读(214)  评论(0)    收藏  举报