[HIHO1328]逃离迷宫（bfs，位压）

题目链接：http://hihocoder.com/problemset/problem/1328

这个题bfs到时候不止要存当前的坐标，还要存当前有哪几把钥匙。因为5把钥匙，所以可以直接用位来存，这样也可以解决一个房间里有好几把钥匙的情况。

还有就是走的过程中，一个点重复走多少次的问题，我们用vis(x,y,k)来记录坐标(x,y)的时候拿着钥匙位压后为k的情况，并且初始化成0x7f7f7f，每次更新最短路，假如有重复走并且拥有钥匙情况相同的情况就可以剪掉了。

 /*
 ━━━━━┒ギリギリ♂ eye！
 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%I64d", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 typedef struct Node {
     int x, y;
     int s;
     int k;
     Node() {}
     Node(int xx, int yy, int ss, int kk) : x(xx), y(yy), s(ss), k(kk) {}
 }Node;
 const int dx[] = {,-,,};
 const int dy[] = {,,,-};
 const int maxn = ;
 int n, m, k;
 int sx, sy, ex, ey;
 int vis[maxn][maxn][];
 char G[maxn][maxn];
 char ks[maxn][maxn];
 int ret;
 queue<Node> q;

 bool ok(int x, int y) {
     return x >=  && y >=  && x < n && y < m && G[x][y] != '#';
 }

 bool yes(int x, int y, int k) {
     if(G[x][y] == '.') return ;
     return k & ( << (G[x][y] - 'A'));
 }

 void bfs() {
     while(!q.empty()) q.pop();
     q.push(Node(sx, sy, , ));
     vis[sx][sy][] = ;
     while(!q.empty()) {
         Node t = q.front(); q.pop();
         // printf("%d\n", t.k);
         if(t.x == ex && t.y == ey) {
             ret = min(ret, t.s);
             return;
         }
         Rep(i, ) {
             int x = t.x + dx[i];
             int y = t.y + dy[i];
             if(ok(x, y)) {
                 int ck =  t.k | ks[x][y];
                 if(vis[x][y][ck] > t.s +  && yes(x, y, ck)) {
                     vis[x][y][ck] = t.s + ;
                     q.push(Node(x, y, t.s+, ck));
                 }
             }
         }
     }
 }

 int main() {
     // FRead();
     int x, y;
     while(~scanf("%d%d%d%d%d%d%d",&n,&m,&k,&sx,&sy,&ex,&ey)) {
         Clr(vis, 0x7f7f); Cls(G); Cls(ks); ret = 0x7f7f;
         Rep(i, n) Rs(G[i]);
         Rep(i, k) {
             Rint(x); Rint(y);
             ks[x][y] |= ( << i);
         }
         bfs();
         if(ret != 0x7f7f) printf("%d\n", ret);
         else printf("-1\n");
     }
     RT ;
 }

这种判断重复的方法是不对的，因为会在一个出不去而不停兜圈子，拿不到钥匙并且会超时。

 /*
 ━━━━━┒ギリギリ♂ eye！
 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%I64d", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 typedef struct Node {
     int x, y;
     int s;
     int k;
     Node() {}
     Node(int xx, int yy, int ss) : x(xx), y(yy), s(ss) { k = ; }
     Node(int xx, int yy, int ss, int kk) : x(xx), y(yy), s(ss), k(kk) {}
 }Node;
 const int dx[] = {,-,,};
 const int dy[] = {,,,-};
 const int maxn = ;
 int n, m, k;
 int sx, sy, ex, ey;
 char G[maxn][maxn];
 char ks[maxn][maxn];
 int ret;
 queue<Node> q;

 bool ok(int x, int y) {
     return x >=  && y >=  && x < n && y < m && G[x][y] != '#';
 }

 void bfs() {
     while(!q.empty()) q.pop();
     q.push(Node(sx, sy, , ));
     int wtf = ;
     while(!q.empty()) {
         if(wtf++ == ) return;
         Node t = q.front(); q.pop();
         // printf("%d\n", t.k);
         if(t.x == ex && t.y == ey) {
             ret = min(ret, t.s);
             return;
         }
         Rep(i, ) {
             int x = t.x + dx[i];
             int y = t.y + dy[i];
             if(ok(x, y)) {
                 int cur =  t.k | ks[x][y];
                 if(G[x][y] == '.' ||  (cur & ( << (G[x][y] - 'A')))) {
                     q.push(Node(x, y, t.s+, cur));
                 }
             }
         }
     }
 }

 int main() {
     // FRead();
     int x, y;
     while(~scanf("%d%d%d%d%d%d%d",&n,&m,&k,&sx,&sy,&ex,&ey)) {
         Cls(G); Cls(ks); ret = 0x7f7f7f;
         Rep(i, n) Rs(G[i]);
         Rep(i, k) {
             Rint(x); Rint(y);
             ks[x][y] |= ( << i);
         }
         bfs();
         if(ret == 0x7f7f7f) printf("-1\n");
         else printf("%d\n", ret);
     }
     RT ;
 }