• 牛客小白月赛7 CSL的校园卡


    CSL的校园卡

    思路:

    bfs,用状压表示走过的区域,然后和x1,y1,x2,y2构成所有的状态,然后标记一下就可以了

    代码:

    #pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #pragma GCC optimize(4)
    #include<bits/stdc++.h>
    using namespace std;
    #define fi first
    #define se second
    #define pi acos(-1.0)
    #define LL long long
    //#define mp make_pair
    #define pb push_back
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pii pair<int, int>
    #define piii pair<pii, int>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
    //head
     
    bool vis[(1<<16) + 10][5][5][5][5];
    int dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};
    char s[10][10];
    int n, m;
    struct node {
        int x1, y1, x2, y2, dis, st;
    };
    int get(int x, int y) {
        return (x-1)*m + y - 1;
    }
    int bfs() {
        queue<node> q;
        int st = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if(s[i][j] == 'S' || s[i][j] == 'X') st |= 1<<get(i, j);
            }
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if(s[i][j] == 'S') {
                    q.push(node{i, j, i, j, 0, st});
                    break;
                }
            }
        }
        while(!q.empty()) {
            node now = q.front();
            q.pop();
            if(now.st == (1<<n*m) - 1) return now.dis;
            if(!vis[now.st][now.x1][now.y1][now.x2][now.y2]) vis[now.st][now.x1][now.y1][now.x2][now.y2] = true;
            else continue;
            for (int i = 0; i < 4; i++) {
                for (int j = 0; j < 4; j++) {
                    int x = now.x1 + dir[i][0];
                    int y = now.y1 + dir[i][1];
                    int xx = now.x2 + dir[j][0];
                    int yy = now.y2 + dir[j][1];
                    if(1 <= x && x <= n && 1 <= y && y <= m && 1 <= xx && xx <= n && 1 <= yy && yy <= m && s[x][y] != 'X' && s[xx][yy] != 'X') {
                        q.push(node{x, y, xx, yy, now.dis+1, now.st|(1<<get(x, y))|(1<<get(xx, yy))});
                    }
                }
            }
        }
        return 0;
    }
    int main() {
        scanf("%d %d", &n, &m);
        for (int i = 1; i <= n; i++) {
            scanf("%s", s[i]+1);
        }
        printf("%d
    ", bfs());
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/widsom/p/9704098.html
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