其实这题可以直接二进制状压做，1表示黑棋，0表示白棋，另外记录下2个空点的位置就行了。 具体看代码（冗长）：
#include #include #include #include #include #include #include #include #include #include #include #define rep(i,m,n) for(int i=m;i<=n;++i) #define dop(i,m,n) for(int i=m;i>=n;--i) #define lowbit(x) (x&(-x)) #define ll long long #define INF 2147483647 #define re register #define Open(s) freopen(s".in","r",stdin);freopen(s".out","w",stdout); #define Close fclose(stdin);fclose(stdout); #define pause system("pause"); using namespace std; inline int read(){ int s = 0, w = 1; char ch = getchar(); while(ch < '0' || ch > '9'){if(ch == '-')w = -1;ch = getchar();} while(ch >= '0' && ch <= '9') s = s * 10 + ch - '0',ch = getchar(); return s * w; } char a[6][6]; struct Node{ int x[2], y[2], s, time, who; //x,y数组记录2个空点的位置，s表示当前的状态（二进制），who表示到谁走了 }start, begin, now, tmp; int getHash(){ //棋盘转二进制状态 int ans = 0; rep(i, 1, 4) rep(j, 1, 4) if(a[i][j] == 'B') ans |= (1 << ( ((i - 1) << 2) + j - 1) ); return ans; } queue q; bool Win(Node x){ //判赢函数 int s = x.s; //判断有没有黑点四子连棋 if( s&1&&s&(1<<4)&&s&(1<<8)&&s&(1<<12) || s&(1<<1)&&s&(1<<5)&&s&(1<<9)&&s&(1<<13) || s&(1<<2)&&s&(1<<6)&&s&(1<<10)&&s&(1<<14) || s&(1<<3)&&s&(1<<7)&&s&(1<<11)&&s&(1<<15) || s&1&&s&(1<<1)&&s&(1<<2)&&s&(1<<3) || s&(1<<4)&&s&(1<<5)&&s&(1<<6)&&s&(1<<7) || s&(1<<8)&&s&(1<<9)&&s&(1<<10)&&s&(1<<11) || s&(1<<12)&&s&(1<<13)&&s&(1<<14)&&s&(1<<15) || s&1&&s&(1<<5)&&s&(1<<10)&&s&(1<<15) || s&(1<<3)&&s&(1<<6)&&s&(1<<9)&&s&(1<<12)) return true; //把空点的位置赋值为1 s|=1<<(((x.x[0]-1)<<2)+x.y[0]-1); s|=1<<(((x.x[1]-1)<<2)+x.y[1]-1); //rep(i, 1, 4){ rep(j, 1, 4) printf("%d ", (s>>(((i-1)<<2)+j-1))&1); printf("\n"); } //判断白点有没有四子连棋 if( (s^1)&1&&(s^(1<<4))&(1<<4)&&(s^(1<<8))&(1<<8)&&(s^(1<<12))&(1<<12) || (s^(1<<1))&(1<<1)&&(s^(1<<5))&(1<<5)&&(s^(1<<9))&(1<<9)&&(s^(1<<13))&(1<<13) || (s^(1<<2))&(1<<2)&&(s^(1<<6))&(1<<6)&&(s^(1<<10))&(1<<10)&&(s^(1<<14))&(1<<14) || (s^(1<<3))&(1<<3)&&(s^(1<<7))&(1<<7)&&(s^(1<<11))&(1<<11)&&(s^(1<<15))&(1<<15) || (s^1)&&(s^(1<<1))&(1<<1)&&(s^(1<<2))&(1<<2)&&(s^(1<<3))&(1<<3) || (s^(1<<4))&(1<<4)&&(s^(1<<5))&(1<<5)&&(s^(1<<6))&(1<<6)&&(s^(1<<7))&(1<<7) || (s^(1<<8))&(1<<8)&&(s^(1<<9))&(1<<9)&&(s^(1<<10))&(1<<10)&&(s^(1<<11))&(1<<11) || (s^(1<<12))&(1<<12)&&(s^(1<<13))&(1<<13)&&(s^(1<<14))&(1<<14)&&(s^(1<<15))&(1<<15) || (s^1)&(1)&&(s^(1<<5))&(1<<5)&&(s^(1<<10))&(1<<10)&&(s^(1<<15))&(1<<15) || (s^(1<<3))&(1<<3)&&(s^(1<<6))&(1<<6)&&(s^(1<<9))&(1<<9)&&(s^(1<<12))&(1<<12)) return true; return false; } int l[] = { 233, -1, 1, 0, 0 }, r[] = { 666, 0, 0, -1, 1 }, id, X, Y, kong, v[20000010]; int getHash(Node x){ //不要hack我 return x.s + x.x[0] * 10007 + x.x[1] * 107 + x.y[1] * 677 + x.y[0] * 97; } int main(){ rep(i, 1, 4) rep(j, 1, 4){ cin>>a[i][j]; if(a[i][j] == 'O') //记录空点位置 if(!start.x[0]) start.x[0] = i, start.y[0] = j; else start.x[1] = i, start.y[1] = j; } start.s = getHash(); start.time = 0; begin = start; start.who = 0; begin.who = 1; q.push(start); q.push(begin); v[getHash(begin)] = v[getHash(start)] = 1; while(!q.empty()){ //bfs Node now = q.front(); q.pop(); rep(i, 0, 1){ rep(j, 1, 4){ X = now.x[i] + l[j]; Y = now.y[i] + r[j]; if(X <= 0 || X > 4 || Y <= 0 || Y > 4 || X == now.x[i^1] && Y == now.y[i^1]) continue; //不能出界，也不能走到另外一个空点 id = ((X - 1) << 2) + Y - 1; kong = ((now.x[i] - 1) << 2) + now.y[i] - 1; if(((now.s >> id) & 1) == now.who){ //如果是到这个棋走 tmp = now; tmp.s &= ~(1 << id); //交换 tmp.s |= tmp.who << kong; //位置 tmp.x[i] = X; tmp.y[i] = Y; if(v[getHash(tmp)]) continue; v[getHash(tmp)] = 1; tmp.time++; tmp.who^=1; if(Win(tmp)){ printf("%d\n", tmp.time); return 0; } q.push(tmp); } } } } return 0; }
转载于:https://www.cnblogs.com/Qihoo360/p/9468315.html
