SGU536 Berland Chess

棋盘上白子只有一个国王  黑子给出

各子遵从国际象棋的走法

黑子不动，白子不能走进黑子的攻击范围以内

问白字能不能吃掉所有的黑子

直接搜索就好了，各子状态用二进制表示

不过每个子被吃之后攻击范围会改变

所以重点是预处理每种剩余棋子状态的攻击范围

比较麻烦，注意白子吃掉一颗子之后所在的位置也可能是危险位置

//#pragma comment(linker, "/STACK:102400000,102400000")//HEAD#include <cstdio>#include <ctime>#include <cstdlib>#include <cstring>#include <queue>#include <string>#include <set>#include <stack>#include <map>#include <cmath>#include <vector>#include <iostream>#include <algorithm>using namespace std;//LOOP#define FF(i, a, b) for(int i = (a); i < (b); ++i)#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FED(i, b, a) for(int i = (b); i>= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))#define CPY(a, b) memcpy(a, b, sizeof(a))//STL#define PB push_back//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RS(s) scanf("%s", s)//OUTPUT#define WI(n) printf("%d\n", n)#define WS(s) printf("%s\n", s)typedef long long LL;typedef unsigned long long ULL;typedef vector <int> VI;const int INF = 100000000;const double eps = 1e-10;const int maxn = 16;const LL MOD = 1e9 + 7;const int k = 1, b = 2, r = 3;                        ////表示棋子种类int n, m, sx, sy, cnt, all;bool danger[maxn][maxn][1 << 15], vis[maxn][maxn][1 << 15];int dirk[][2] = {{-2, 1}, {-1, 2}, {1, 2}, {2, 1},                {2, -1}, {1, -2}, {-1, -2}, {-2, -1}};int dir[][2] = {{-1, -1}, {-1, 1}, {1, 1}, {1, -1},                {-1, 0}, {1, 0}, {0, -1}, {0, 1}};struct node{    int x, y, type;}pie[maxn];struct state{    short int x, y, key;    int step;    state(int a, int b, int c, int d) : x(a), y(b), key(c), step(d){}};char mat[maxn][maxn];int num[maxn][maxn];int bfs(){    int tot = all - 1;    CLR(vis, 0);    vis[sx][sy][0] = 1;    queue<state> Q;    state t(sx, sy, 0, 0);    Q.push(t);    while (!Q.empty())    {        t = Q.front(); Q.pop();        if (t.key == tot)            return t.step;        REP(i, 8)        {            int nx = t.x + dir[i][0], ny = t.y + dir[i][1], key = t.key;            if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;            if (num[nx][ny] != -1)              ///若这个位置是棋子                key |= (1 << num[nx][ny]);            if (danger[nx][ny][key])    continue;     ///key要是更新后的状态            if (!vis[nx][ny][key])            {                vis[nx][ny][key] = 1;                state heh(nx, ny, key, t.step + 1);                Q.push(heh);            }        }    }    return -1;}void solve(){    all = 1 << cnt;    CLR(danger, 0);    int x, y;    REP(i, cnt)    {        int s0 = 1 << i;        if (pie[i].type == k)        {            REP(j, 8)            {                x = pie[i].x + dirk[j][0], y = pie[i].y + dirk[j][1];                if (x < 0 || x >= n || y < 0 || y >= m)   continue;                REP(s0, all)                    if ((s0 & (1 << i)) == 0)                        danger[x][y][s0] = 1;            }        }        else if (pie[i].type == b)        {            REP(j, 4)            {                int st = 1;                while (1)                {                    x = pie[i].x + dir[j][0] * st, y = pie[i].y + dir[j][1] * st;                    if (x < 0 || x >= n || y < 0 || y >= m) break;                    REP(s0, all)                        if ((s0 & (1 << i)) == 0)                            danger[x][y][s0] = 1;                    st++;                    if (mat[x][y] != '.')   break;              ///这里注意，先是标记这个点，再结束                }            }        }        else if (pie[i].type == r)        {            FF(j, 4, 8)            {                int st = 1;                while (1)                {                    x = pie[i].x + dir[j][0] * st, y = pie[i].y + dir[j][1] * st;                    if (x < 0 || x >= n || y < 0 || y >= m)  break;                    REP(s0, all)                        if ((s0 & (1 << i)) == 0)                            danger[x][y][s0] = 1;                    st++;                    if (mat[x][y] !=  '.')  break;                }            }        }    }    WI(bfs());}int main(){    while (~RII(n, m))    {        cnt = 0;        CLR(num, -1);        REP(i, n)        {            RS(mat[i]);            REP(j, m)            {                if (mat[i][j] == '*')   sx = i, sy = j, mat[i][j] = '.';                else if (mat[i][j] == 'K') num[i][j] = cnt, pie[cnt].x = i, pie[cnt].y = j, pie[cnt].type = k, cnt++;                else if (mat[i][j] == 'B') num[i][j] = cnt, pie[cnt].x = i, pie[cnt].y = j, pie[cnt].type = b, cnt++;                else if (mat[i][j] == 'R') num[i][j] = cnt, pie[cnt].x = i, pie[cnt].y = j, pie[cnt].type = r, cnt++;            }        }        solve();    }}