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;
}