POJ 3322 Bloxorz(算竞进阶习题)

bfs

标准广搜题，主要是把每一步可能的坐标都先预处理出来，会好写很多

每个状态对应三个限制条件，x坐标、y坐标、lie=0表示直立在(x,y)，lie=1表示横着躺，左半边在(x,y)，lie=2表示竖着躺，上半边在(x,y)

#include <bits/stdc++.h>

#define INF 0x3f3f3f3f

using namespace std;

typedef long long ll;

inline int lowbit(int x){ return x & (-x); }

inline int read(){

    int X = 0, w = 0; char ch = 0;

    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }

    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();

    return w ? -X : X;

}

inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }

inline int lcm(int a, int b){ return a / gcd(a, b) * b; }

template<typename T>

inline T max(T x, T y, T z){ return max(max(x, y), z); }

template<typename T>

inline T min(T x, T y, T z){ return min(min(x, y), z); }

template<typename A, typename B, typename C>

inline A fpow(A x, B p, C lyd){

    A ans = 1;

    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;

    return ans;

}

char g[505][505];

struct rec { int x, y, lie; } st, ed;

int r, c;

const int dx[] = {0, 0, 1, -1};

const int dy[] = {1, -1, 0, 0};

int f[505][505][3];

const int nextx[3][4] = {{-2, 1, 0, 0}, {-1, 1, 0, 0}, {-1, 2, 0, 0}};

const int nexty[3][4] = {{0, 0, -2, 1}, {0, 0, -1, 2}, {0, 0, -1, 1}};

const int nextlie[3][4] = {{2, 2, 1, 1}, {1, 1, 0, 0}, {0, 0, 2, 2}};

bool inArea(int x, int y){

    return x >= 0 && x < r && y >= 0 && y < c;

}

bool valid(const rec &next){

    int x = next.x, y = next.y, lie = next.lie;

    if(!inArea(x, y)) return false;

    if(g[x][y] == '#') return false;

    if(lie == 0 && g[x][y] == 'E') return false;

    if(lie == 1 && g[x][y + 1] == '#') return false;

    if(lie == 2 && g[x + 1][y] == '#') return false;

    return true;

}

void init(){

    for(int i = 0; i < r; i ++){

        for(int j = 0; j < c; j ++){

            if(g[i][j] == 'O')

                ed.x = i, ed.y = j, ed.lie = 0, g[i][j] = '.';

            if(g[i][j] == 'X'){

                bool flag = false;

                for(int d = 0; d < 4; d ++){

                    int nx = i + dx[d];

                    int ny = j + dy[d];

                    if(inArea(nx, ny) && g[nx][ny] == 'X'){

                        st.x = min(i, nx), st.y = min(j, ny);

                        st.lie = d < 2 ? 1 : 2;

                        g[nx][ny] = g[i][j] = '.';

                        flag = true;

                    }

                }

                if(!flag){

                    st.x = i, st.y = j, st.lie = 0;

                    g[i][j] = '.';

                }

            }

        }

    }

}

int bfs(){

    memset(f, -1, sizeof f);

    queue<rec> q;

    f[st.x][st.y][st.lie] = 0;

    q.push(st);

    while(!q.empty()){

        rec cur = q.front(); q.pop();

        //cout << cur.x << " " << cur.y << endl;

        for(int i = 0; i < 4; i ++){

            rec next;

            next.x = cur.x + nextx[cur.lie][i];

            next.y = cur.y + nexty[cur.lie][i];

            next.lie = nextlie[cur.lie][i];

            if(valid(next) && f[next.x][next.y][next.lie] == -1){

                f[next.x][next.y][next.lie] = f[cur.x][cur.y][cur.lie] + 1;

                if(next.x == ed.x && next.y == ed.y && next.lie == ed.lie)

                    return f[next.x][next.y][next.lie];

                q.push(next);

            }

        }

    }

    return -1;

}

int main(){

    while(scanf("%d%d", &r, &c) != EOF && r && c){

        for(int i = 0; i < r; i ++) scanf("%s", g[i]);

        init();

        int ans = bfs();

        if(ans == -1) printf("Impossible\n");

        else printf("%d\n", ans);

    }

    return 0;

}