zoukankan      html  css  js  c++  java
  • hdu 4634 Swipe Bo bfs+状态压缩

    题目链接

    状态压缩记录当前拿到了哪些钥匙, 然后暴力搜索。

    搞了好几个小时, 一开始也不知道哪里错了, 最后A了也不知道一开始哪里有问题。

    #include <iostream>
    #include <vector>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <map>
    #include <set>
    #include <string>
    #include <queue>
    #include <stack>
    #include <bitset>
    using namespace std;
    #define pb(x) push_back(x)
    #define ll long long
    #define mk(x, y) make_pair(x, y)
    #define lson l, m, rt<<1
    #define mem(a) memset(a, 0, sizeof(a))
    #define rson m+1, r, rt<<1|1
    #define mem1(a) memset(a, -1, sizeof(a))
    #define mem2(a) memset(a, 0x3f, sizeof(a))
    #define rep(i, n, a) for(int i = a; i<n; i++)
    #define fi first
    #define se second
    typedef pair<int, int> pll;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int mod = 1e9+7;
    const int inf = 1061109567;
    const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
    const int maxn = 201;
    int n, m, cnt, g[maxn][maxn];
    bool vis[maxn][maxn][1<<7][4], used[maxn][maxn][1<<7];
    char s[maxn][maxn];
    struct node
    {
        int x, y, step, key;
        node(){}
        node(int _x, int _y, int _step, int _key): x(_x), y(_y), step(_step), key(_key){}
    };
    queue <node> q;
    int check(int x, int y) {
        if(x>=0&&x<n&&y>=0&&y<m)
            return 1;
        return 0;
    }
    int judge(int x, int y) {
        if(x == 1)
            return 0;
        if(x == -1)
            return 1;
        if(y == 1)
            return 2;
        if(y == -1)
            return 3;
    }
    void bfs(int x, int y) {
        mem(vis);
        mem(used);
        while(!q.empty())
            q.pop();
        used[x][y][0] = 1;
        q.push(node(x, y, 0, 0));
        while(!q.empty()) {
            node temp = q.front(); q.pop();
            int x = temp.x, y = temp.y, tmpx, tmpy;
            for(int i = 0; i < 4; i++) {
                int dirx = dir[i][0], diry = dir[i][1];
                tmpx = x, tmpy = y;
                int key = temp.key;
                node tmp = temp;
                if(!check(tmpx+dirx, tmpy+diry) || s[tmpx+dirx][tmpy+diry]=='#')
                    continue;
                while(1) {
                    if(s[tmpx][tmpy] == 'L')
                        dirx = 0, diry = -1;
                    if(s[tmpx][tmpy] == 'R')
                        dirx = 0, diry = 1;
                    if(s[tmpx][tmpy] == 'U')
                        dirx = -1, diry = 0;
                    if(s[tmpx][tmpy] == 'D')
                        dirx = 1, diry = 0;
                    int tmpdir = judge(dirx, diry);
                    if(vis[tmpx][tmpy][tmp.key][tmpdir])
                        break;
                    vis[tmpx][tmpy][tmp.key][tmpdir] = 1;
                    if(s[tmpx][tmpy] == 'E' && key == (1<<cnt)-1) {
                        printf("%d
    ", temp.step+1);
                        return ;
                    }
                    if(s[tmpx][tmpy] == 'K')
                        key |= g[tmpx][tmpy];
                    if(check(tmpx+dirx, tmpy+diry)) {
                        if(s[tmpx+dirx][tmpy+diry] == '#') {
                            tmp.x = tmpx;
                            tmp.y = tmpy;
                            tmp.step++;
                            tmp.key = key;
                            q.push(tmp);
                            used[tmpx][tmpy][tmp.key] = 1;
                            break;
                        } else {
                            tmpx += dirx;
                            tmpy += diry;
                        }
                    } else {
                        break;
                    }
                }
            }
        }
        puts("-1");
        return ;
    }
    void solve() {
        cnt = 0;
        int x, y;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(s[i][j] == 'S')
                    x = i, y = j;
                if(s[i][j] == 'K') {
                    g[i][j] = 1<<cnt;
                    cnt++;
                }
            }
        }
        bfs(x, y);
    }
    void read() {
        for(int i = 0; i < n; i++)
            scanf("%s", s[i]);
    }
    int main()
    {
        while(scanf("%d%d", &n, &m)!=EOF) {
            read();
            solve();
        }
        return 0;
    }
  • 相关阅读:
    CVE-2017-10271
    [GKCTF2020]cve版签到
    [GXYCTF2019]禁止套娃 无参数RCE
    [护网杯 2018]easy_tornado
    记两道xctf上的web进阶区 反序列化
    msf卸载win defender
    Cron表达式详解
    Linux ifconfig只有lo没有别的网络的问题
    记一道文件上传
    【解决】手机安卓已经导入burp证书,但仍提示此证书并非来自被信任的机构
  • 原文地址:https://www.cnblogs.com/yohaha/p/5289513.html
Copyright © 2011-2022 走看看