拯救大兵瑞恩 HDU - 4845（状压bfs || 分层最短路）

Posted 2021-01-10 wtsruvf

1、状压bfs

这个状压体现在key上  我i们用把key状压一下  就能记录到一个点时 已经拥有的key的种类

ban[x1][y1][x2][y1]记录两个点之间的状态 是门 还是墙 还是啥都没有

inc[x][y]记录这个点所存储的钥匙  （可能不止一个 所以要用二进制）

vis[x][y][key]  标记当前点 在拥有的钥匙种类为key时是否走过

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define pd(a) printf("%d
", a);
#define plld(a) printf("%lld
", a);
#define pc(a) printf("%c
", a);
#define ps(a) printf("%s
", a);
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 20, INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;
int n, m, p, k, s;
int ban[maxn][maxn][maxn][maxn], inc[maxn][maxn];
int vis[maxn][maxn][1 << 10 + 1];
int dis[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

struct node
{
    int x, y, key, d;
    node(int x, int y, int key, int d) : x(x), y(y), key(key), d(d) {}
};

int bfs()
{
    queue<node> Q;
    mem(vis, 0);
    Q.push(node(1, 1, 0 | inc[1][1], 0));
    vis[1][1][0 | inc[1][1]] = 1;
    while(!Q.empty())
    {
        node is = Q.front(); Q.pop();
        for(int i = 0; i < 4; i++)
        {
            node it(0, 0, 0, 0);
            it.x = is.x + dis[i][0];
            it.y = is.y + dis[i][1];
            if(it.x < 1 || it.x > n || it.y < 1 || it.y > m)
                continue;
            if(!ban[it.x][it.y][is.x][is.y])
                continue;
            if(ban[it.x][it.y][is.x][is.y] == -1 || is.key & (1 << (ban[it.x][it.y][is.x][is.y] -1)))  //如果下一个点与当前点没有墙或门 或者到当前点时已经有了进入下一个点的钥匙
            {
                if(!inc[it.x][it.y]) it.key = is.key;  //如果下一个点没有存储的钥匙 则直接把当前钥匙传递过去
                else it.key = is.key | inc[it.x][it.y]; //如果有存储的钥匙 把存储的钥匙加上即可
                if(!vis[it.x][it.y][it.key])
                {
                    vis[it.x][it.y][it.key] = 1;
                    it.d = is.d + 1;
                    Q.push(it);
                    if(it.x == n && it.y == m)
                        return it.d;
                }
            }
        }
    }
    return -1;
}


int main()
{
    while(cin >> n >> m >> p >> k)
    {
        int x1, y1, x2, y2, G, Q;
        mem(ban, -1);
        mem(inc, 0);
        for(int i = 1; i <= k; i++)
        {
            cin >> x1 >> y1 >> x2 >> y2 >> G;
            ban[x1][y1][x2][y2] = ban[x2][y2][x1][y1] = G;
        }
        cin >> s;
        for(int i = 1; i <= s; i++)
        {
            cin >> x1 >> y1 >> Q;
            inc[x1][y1] |= (1 << (Q - 1));
        }
        cout << bfs() << endl;

    }

    return 0;
}