hdu 5094 Maze

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题目链接

分析：
把十把钥匙的状态进行二进制压缩，然后就是bfs了。
坑点就是两个点之间可能有很多门，一个地方可能有多把钥匙。之前一直WA，我看了题解才做对＝＝

代码：

#include <iostream>#include <cstdio>#include <vector>#include <cstring>#include <stack>#include <queue>using namespace std;int N, M, P;const int dx[] = {0, 1, 0, -1};const int dy[] = {1, 0, -1, 0};const int INF = 0x3f3f3f3f;int g[55][55][55][55];int mp[55][55];bool vis[55][55][1 << 11];struct Node {    int x, y, t, d;};bool valid(int x, int y) {    return (x >= 1 && x <= N && y >= 1 && y <= M);}int bfs() {    memset(vis, false, sizeof(vis));    queue<Node> q;    q.push((Node){1, 1, mp[1][1], 0});    vis[1][1][mp[1][1]] = true;    while (!q.empty()) {        Node x = q.front(); q.pop();//        cout<<x.x<<" "<<x.y<<" "<<x.t<<endl;        if (x.x == N && x.y == M) return x.d;        for (int k = 0; k < 4; k ++) {            int tmpx = x.x + dx[k];            int tmpy = x.y + dy[k];            if (!valid(tmpx, tmpy)) continue;            int key = g[x.x][x.y][tmpx][tmpy];            if (key == INF) continue;            if ((key & x.t) != key) continue;            int tmpt = x.t | mp[tmpx][tmpy];            if (vis[tmpx][tmpy][tmpt]) continue;            vis[tmpx][tmpy][tmpt] = true;            q.push((Node){tmpx, tmpy, tmpt, x.d + 1});        }    }    return -1;}int main(int argc, char const *argv[]) {    while(~scanf("%d%d%d", &N, &M, &P)) {        memset(g, 0, sizeof(g));        memset(mp, 0, sizeof(mp));        int k;        scanf("%d", &k);        for (int i = 0; i < k; i ++) {            int a, b, c, d, e;            scanf("%d%d%d%d%d", &a, &b, &c, &d, &e);            g[a][b][c][d] |= 1 << e;            g[c][d][a][b] |= 1 << e;            if (!e) g[a][b][c][d] = g[c][d][a][b] = INF;        }        scanf("%d", &k);        for (int i = 0; i < k; i ++) {            int a, b, c;            scanf("%d%d%d", &a, &b, &c);            mp[a][b] |= 1 << c;        }        cout<<bfs()<<endl;    }    return 0;}

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