Maze

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)

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 x_i1, y_i1, x_i2, y_i2, g_i; when g_i >=1, represents there is a gate of type gi between location (x_i1, y_i1) and (x_i2,
y_i2); when g_i = 0, represents there is a wall between location (x_i1, y_i1) and (x_i2, y_i2), ( | x_i1 - x_i2 | + | y_i1 - y_i2 |=1, 0<= g_i <=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 x_i1, y_i1 and q_i respectively. That means the key type of q_i locates on location (x_i1, y_i1), (1<= q_i<=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

把 map 数组初始化为 -1 了 ，测试样例都过不了，调试了2个多小时才发现 ，真是越来越都比了。

还要注意：一个地方可能多多个不同类型的钥匙，坑啊

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define maxn 55
using namespace std;

int vis[maxn][maxn][1 << 11];
int map[maxn][maxn][maxn][maxn];
int keynum[maxn][maxn][11];//记录这个地点有哪几种钥匙
int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
int n, m, p, s, k;//p代表有几种钥匙。

struct node{
    int x, y ,step, key;
    friend bool operator  < (node a , node b)
    {
        return a.step > b.step;
    }
};

int check(node a, node b)
{
    if(b.x <= 0 || b.x > n || b.y <= 0 || b.y > m)
        return 0;
    if(map[a.x][a.y][b.x][b.y] == 0)
        return 0;
    return 1;
}

int BFS(){
    priority_queue<node>q;
    node now, next;
    now.x = 1;
    now.y = 1;
    now.step = 0;
    now.key = 0;
    for(int j = 1; j <= 10; ++j){// 起点有钥匙
        if(keynum[now.x][now.y][j])
            now.key = now.key | (1 << j);
    }
    q.push(now);
    vis[now.x][now.y][now.key] = 1;
    while(!q.empty()){
        now = q.top();
        q.pop();
        if(now.x == n && now.y == m){
            return now.step;
        }
        for(int i = 0; i < 4; ++i){
            next.x = now.x + dir[i][0];
            next.y = now.y + dir[i][1];
            next.step = now.step + 1;
            //printf("--%d\n", next.step);
            next.key = now.key;
            if(check(now, next)){//now 和 next 之间没有墙
                if(map[now.x][now.y][next.x][next.y] > 0){ //now 和 next 中间有门
                    if(next.key & (1 << map[now.x][now.y][next.x][next.y])){//有这个门的钥匙，能通过这个门到达next
                        for(int j = 1; j <= 10; ++j){ //判断点next是不是有钥匙
                            if(keynum[next.x][next.y][j])
                                next.key = next.key | (1 << j);
                        }
                        if(!vis[next.x][next.y][next.key]){
                            vis[next.x][next.y][next.key] = 1;
                            q.push(next);
                        }
                    }
                }
                else {
                    for(int j = 1; j <= 10; ++j){//判断点next是不是有钥匙
                        if(keynum[next.x][next.y][j]){
                            next.key = next.key | (1 << j);
                        }
                    }
                    if(!vis[next.x][next.y][next.key]){
                            vis[next.x][next.y][next.key] = 1;
                            q.push(next);
                    }
                }
            }
        }
    }
    return -1;
}

int main (){
    while(scanf("%d%d%d", &n, &m, &p) != EOF){
        scanf("%d", &k);
        memset(map, -1, sizeof(map));
        memset(vis, 0, sizeof(vis));
        memset(keynum, 0, sizeof(keynum));
        while(k--){
            int x1, y1, x2, y2, t;
            scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &t);
            map[x1][y1][x2][y2] = t;
            map[x2][y2][x1][y1] = t;
        }
        scanf("%d", &s);
        while(s--){
            int x, y, t;
            scanf("%d%d%d", &x, &y, &t);
            keynum[x][y][t] = 1;
        }
        printf("%d\n", BFS());
    }
    return 0;
}

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