Traffic Jam
| Source : CTU Open 2003 | |||
| Time limit : 5 sec | Memory limit : 32 M | ||
Submitted : 14, Accepted : 13
Traffic jam is a real nightmare of all drivers. Nobody likes to be stuck in the overfilled streets, when the cars move very slowly, if they even move at all. Professional drivers face traffic jams quite often. Truck drivers of ACM are not an exception. Can you help drivers to find the way out of the traffic jam?
We can model a small (but complicated) traffic jam on a 6 × 6grid of squares. Vehicles (cars and trucks) are scattered over the grid at integer locations, as shown below. Both types of vehicles are 1 square wide. Cars are 2 squares long, and trucks are 3 squares long. Vehicles may be oriented either horizontally (East-West) or vertically (North-South) relative to the grid.
Vehicles cannot move through each other, cannot turn, and cannot move over the edge of the grid. They can move in their direction (horizontally-oriented vehicles cannot move vertically and vice versa), as long as they are not blocked by another vehicle or by the edge of the grid. Only one vehicle may move in a single step, but it may move by as many squares at a time as possible, providing there is enough empty space.
Our goal is to move vehicles back and forth until a particular horizontally-oriented vehicle (your own car -- the black one on the picture above) leaves the rightmost (eastern-most) edge of the grid, where it is considered to have escaped the traffic jam. You are to write a program that will find a solution requiring the minimum possible number of moves.
Input
The input file will consist of one or more input scenarios. Each scenario begins with a single integer n, 1 ≤ n ≤ 10, giving the number of vehicles in the scenario. Then there will be 6 lines of input, each containing 6 characters. Each character is either a dot (".) representing an empty square, or a lowercase letter representing a vehicle. Your own vehicle is always oriented horizontally and represented by "x characters. The other vehicles use the letters sequentially, beginning with "a.
The last scenario will be followed by a line containing a single zero.
Output
For each scenario, output a single line with the statement "Scenario #K requires X moves., where K is the number of the scenario (starting with 1) and X is the minimum number of moves required to escape the the traffic jam with the particular car.
If it is not possible to escape, output the sentence "You are trapped in scenario #K. instead.
Sample Input
8 aa...b c..d.b cxxd.b c..d.. e...ff e.ggg. 8 abbc.. a..c.. axxc.. ..gddd ..g..e ..fffe 0Sample Output
Scenario #1 requires 8 moves. Scenario #2 requires 25 moves.
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<stack>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
#define MOD 89991 //自己根据需要改大小
#define LL long long
int Move[2][4] = { { 0,1,0,-1},{1,0,-1,0} };
int n , Cas;
char maze[7][7];
bool b[26];
inline bool inRange(int r,int c)
{
return 0 <= r && r < 6 && 0 <= c && c < 6;
}
struct Car
{
int r , c;
int dir , len;
};
struct State
{
bool vis[6][6];
Car car[10];
State * next;
int step;
}vis[MOD] , *first[MOD] , start;
int sz;
int hashcode(const State & s)
{
LL ret = 0;
for (int i = 0 ; i < n ; ++i)
{
ret += 1313L*((LL)s.car[i].r*131313^s.car[i].c*13131313)<<2;
ret %= MOD;
}
return ret;
}
bool equals(const State & s1 , const State & s2)
{
for (int i = 0 ; i < n ; ++i)
{
if (s1.car[i].r!=s2.car[i].r || s1.car[i].c!=s2.car[i].c) return false;
}
return true;
}
bool in_vis(const State & s)
{
int k = hashcode(s);
State * p = first[k];
while (p)
{
if (equals(s,*p)) return true;
p = p->next;
}
return false;
}
void add(const State & s)
{
vis[sz] = s;
int k = hashcode(s);
vis[sz].next = first[k];
first[k] = &vis[sz];
++sz;
}
void init()
{
memset(first,0,sizeof(first));
memset(vis,0,sizeof(vis));
sz = 0;
}
void input()
{
for (int i = 0 ; i < 6 ; ++i)
scanf("%s",maze[i]);
memset(start.vis,0x3f,sizeof(start.vis));
memset(b,0,sizeof(b));
int size = 1;
start.step = 0;
for (int i = 0 ; i < 6 ; ++i)
{
for (int j = 0 ; j < 6 ; ++j)
{
if (maze[i][j]=='.') continue;
start.vis[i][j] = false;
if (b[maze[i][j]-'a']) continue;
b[maze[i][j]-'a'] = true;
int Index = maze[i][j]=='x' ? 0 : size++;
Car & veh = start.car[Index];
veh.len = 1;
veh.r = i , veh.c = j;
while (inRange(i,j+veh.len) && maze[i][j]==maze[i][j+veh.len])
{
start.vis[i][j+veh.len] = false;
++veh.len;
}
if (veh.len != 1)
{
veh.dir = 0;
continue;
}
while (inRange(i+veh.len,j) && maze[i][j]==maze[i+veh.len][j])
{
start.vis[i+veh.len][j] = false;
++veh.len;
}
veh.dir = 1;
}
}
add(start);
}
bool finish(const State & s)
{
int dir = s.car[0].dir , len = s.car[0].len , r = s.car[0].r , c = s.car[0].c;
int i = r+Move[0][dir]*len , j = c+Move[1][dir]*len;
bool ok = true;
while (inRange(i,j))
{
if (!s.vis[i][j])
{
ok = false;
break;
}
i += Move[0][dir];
j += Move[1][dir];
}
return ok;
}
void solve()
{
if (start.car[0].dir==1)
{
printf("You are trapped in scenario #%d.\n",Cas);
return;
}
queue<State> q;
q.push(start);
while (q.size())
{
State tmp = q.front(); q.pop();
for (int i = 0 ; i < n ; ++i)
{
State now = tmp;
++now.step;
int len = now.car[i].len , d = now.car[i].dir;
int & r1 = now.car[i].r;
int & c1 = now.car[i].c;
int r2 = r1+(len-1)*Move[0][d];
int c2 = c1+(len-1)*Move[1][d];
while (inRange(r2,c2))
{
r2 += Move[0][d];
c2 += Move[1][d];
if (!inRange(r2,c2) || !now.vis[r2][c2]) break;
now.vis[r1][c1] = true;
r1 += Move[0][d];
c1 += Move[1][d];
now.vis[r2][c2] = false;
if (!in_vis(now))
{
if (i > 0 && finish(now))
{
printf("Scenario #%d requires %d moves.\n",Cas,now.step+1);
return;
}
q.push(now);
add(now);
}
}
now = tmp;
r1 = now.car[i].r;
c1 = now.car[i].c;
r2 = r1+(len-1)*Move[0][d];
c2 = c1+(len-1)*Move[1][d];
d = (d+2)%4;
++now.step;
while (inRange(r1,c1))
{
r1 += Move[0][d];
c1 += Move[1][d];
if (!inRange(r1,c1) || !now.vis[r1][c1]) break;
now.vis[r1][c1] = false;
now.vis[r2][c2] = true;
r2 += Move[0][d];
c2 += Move[1][d];
if (!in_vis(now))
{
if (i > 0 && finish(now))
{
printf("Scenario #%d requires %d moves.\n",Cas,now.step+1);
return;
}
q.push(now);
add(now);
}
}
}
}
printf("You are trapped in scenario #%d.\n",Cas);
}
int main()
{
while (scanf("%d",&n),n)
{
++Cas;
init();
input();
solve();
}
}
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