Maze
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 901 Accepted Submission(s): 314
Problem Description
This story happened on the background of Star Trek.
Spock, the deputy captain of Starship Enterprise, fell into Klingon’s trick and was held as prisoner on their mother planet Qo’noS.
The captain of Enterprise, James T. Kirk, had to fly to Qo’noS to rescue his deputy. Fortunately, he stole a map of the maze where Spock was put in exactly.
The maze is a rectangle, which has n rows vertically and m columns horizontally, in another words, that it is divided into n*m locations. An ordered pair (Row No., Column No.) represents a location in the maze. Kirk moves from current location to next costs
1 second. And he is able to move to next location if and only if:
Next location is adjacent to current Kirk’s location on up or down or left or right(4 directions)
Open door is passable, but locked door is not.
Kirk cannot pass a wall
There are p types of doors which are locked by default. A key is only capable of opening the same type of doors. Kirk has to get the key before opening corresponding doors, which wastes little time.
Initial location of Kirk was (1, 1) while Spock was on location of (n, m). Your task is to help Kirk find Spock as soon as possible.
Input
The input contains many test cases.
Each test case consists of several lines. Three integers are in the first line, which represent n, m and p respectively (1<= n, m <=50, 0<= p <=10).
Only one integer k is listed in the second line, means the sum number of gates and walls, (0<= k <=500).
There are 5 integers in the following k lines, represents xi1, yi1, xi2, yi2, gi; when gi >=1, represents there is a gate of type gi between location (xi1, yi1) and (xi2,
yi2); when gi = 0, represents there is a wall between location (xi1, yi1) and (xi2, yi2), ( | xi1 - xi2 | + | yi1 - yi2 |=1, 0<= gi <=p
)
Following line is an integer S, represent the total number of keys in maze. (0<= S <=50).
There are three integers in the following S lines, represents xi1, yi1 and qi respectively. That means the key type of qi locates on location (xi1, yi1), (1<= qi<=p).
Output
Output the possible minimal second that Kirk could reach Spock.
If there is no possible plan, output -1.
Sample Input
4 4 9 9 1 2 1 3 2 1 2 2 2 0 2 1 2 2 0 2 1 3 1 0 2 3 3 3 0 2 4 3 4 1 3 2 3 3 0 3 3 4 3 0 4 3 4 4 0 2 2 1 2 4 2 1
Sample Output
14
大致题意：
有一个N*M的地图，图里面有p种门。
下一行输入一个k，代表有k个门 (或墙)。下面k行给出它们所在的位置，样例（x1，x2，y1，y2，op）意义：前四个数值分别代表两个位置的坐标，当op为0时说明两位置之间是一堵墙，当op大于0时说明两位置之间是一扇门且op代表门的型号。
然后输入一个S，代表钥匙的数目，接下来S行每行三个数（x，y，op）代表位置（x，y）有 一把型号op的门 的钥匙。
问你能够从（1，1）出发到达（N，M），若可以输出最小步数，否则输出-1。当然每个点是可以无限走的。
自己就是脑残，忘了输出-1了。 WA了两次。。。
思路：简单BFS + 状态压缩。注意状态压缩处理二进制时，对门和钥匙的编号要减一处理（不要问我为什么）。
注意此题有坑处！！！
一：每个点可能有 很多把 相同型号或者不同型号的钥匙。
二：如果你想用vector存储每个位置的钥匙，我劝你还是放弃吧。
三：起点可能有钥匙，注意初始状态的处理。
说下代码中数组的意义：
1，三维数组vis[ x ][ y ][ State ]存储在位置（x，y）的State状态。
2，三维数组pos[ x ][ y ][ i ]存储位置（x，y）是否有第 i 种钥匙，若有为1，反之为0。
3，四维数组rec[x1][y1][x2][y2]记录两个位置间是否为墙 或 门 或 什么都没有，-1表示什么都没有，-2表示有墙，大于或等于0表示有门 且 相应数值 为门的型号。
4，Map[ x ][ y ] 为 0 时表示位置（x，y）没有钥匙，大于0时说明有钥匙。
有了以上数组就是裸BFS了，详细看我代码。
提醒：可以选用优先队列进行优化，对于这道题优先队列比普通队列快了点。
AC代码：
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 51
using namespace std;
struct Node
{
int x, y, step, key;
friend bool operator < (Node a, Node b)
{
return a.step > b.step;
}
};
int Map[MAXN][MAXN];
int pos[MAXN][MAXN][10];//存储该位置拥有的 钥匙种类
bool vis[MAXN][MAXN][1<<10];//最多10个门
int rec[MAXN][MAXN][MAXN][MAXN];// -2表示两点间有墙 大于或等于0表示两点间有门 -1表示什么都没有
int N, M, p, k;
void getMap()
{
int x, y, x1, y1, x2, y2, op;
memset(Map, 0, sizeof(Map));//0表示该位置什么都没有
memset(rec, -1, sizeof(rec));//-1表示两个位置之间 什么都没有
scanf("%d", &k);
for(int i = 1; i <= k; i++)//k个地方有门 或者 有墙
{
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &op);
if(op == 0)//墙
rec[x1][y1][x2][y2] = rec[x2][y2][x1][y1] = -2;
else//门
rec[x1][y1][x2][y2] = rec[x2][y2][x1][y1] = op-1;
}
memset(pos, 0, sizeof(pos));
int S;//钥匙数目
scanf("%d", &S);
while(S--)
{
scanf("%d%d%d", &x, &y, &op);
Map[x][y] = 1;
if(!pos[x][y][op-1])//该钥匙 还没有
pos[x][y][op-1] = 1;
}
}
bool judge(Node a)//是否越界
{
return a.x >= 1 && a.x <= N && a.y >= 1 && a.y <= M;
}
int wall_or_door(Node a, Node b)//判断两个点间是否有 墙 或有门 或什么都没有
{
return rec[a.x][a.y][b.x][b.y];
}
void BFS(int x, int y)
{
priority_queue<Node> Q;
int move[4][2] = {0,1, 0,-1, 1,0, -1,0};
memset(vis, false, sizeof(vis));
Node now, next;
now.x = x, now.y = y, now.step = 0, now.key = 0;
if(Map[now.x][now.y])//起点可能有钥匙 这里是个坑
{
for(int i = 0; i < 10; i++)
{
if(pos[now.x][now.y][i])
now.key |= (1 << i);
}
}
Q.push(now);
vis[now.x][now.y][now.key] = true;
while(!Q.empty())
{
now = Q.top();
Q.pop();
if(now.x == N && now.y == M)//到达终点
{
printf("%d\n", now.step);
return ;
}
for(int k = 0; k < 4; k++)
{
next.x = now.x + move[k][0];
next.y = now.y + move[k][1];
next.step = now.step + 1;
int t = wall_or_door(now, next);
if(judge(next) && t != -2)//不能越界且中间不能有墙
{
//分有门 和 没有门 来讨论
if(t >= 0)//有门
{
next.key = now.key;
//先看是否有钥匙 要保证能到达目标位置
if(next.key & (1 << t))//有钥匙
{
//判断目标位置是否有钥匙
if(Map[next.x][next.y])//有钥匙 收集钥匙
{
for(int i = 0; i < 10; i++)
{
if(pos[next.x][next.y][i])
next.key |= (1 << i);
}
}
if(!vis[next.x][next.y][next.key])//判断该状态 是否出现过
{
vis[next.x][next.y][next.key] = true;
Q.push(next);
}
}
//没有钥匙这条路目前不能走
}
else//没有门
{
next.key = now.key;
if(Map[next.x][next.y])//目标位置有钥匙
{
for(int i = 0; i < 10; i++)
{
if(pos[next.x][next.y][i])
next.key |= (1 << i);
}
}
if(!vis[next.x][next.y][next.key])
{
vis[next.x][next.y][next.key] = true;
Q.push(next);
}
}
}
}
}
printf("-1\n");
}
int main()
{
while(scanf("%d%d%d", &N, &M, &p) != EOF)
{
getMap();
BFS(1, 1);
}
return 0;
}
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