连连看游戏中的最短路径
连连看游戏是一个比较简单的游戏,两个相同的点可以消除,但前提是两点之间的路径不能有多于两个折点。如何去求解两个点之间的距离呢?我们可以将这里的距离定义为二元组(x,y),x表示多少次转折,y表示路径长度。x值越小距离越短,相同x值的情况下y值越小距离越短。只要使用BFS就可以了,下面就容易写代码了。
#include <iostream> #include <queue> using namespace std; const int N = 20; int inline getx(int x) { return x & 0xFF; } int inline gety(int x) { return (x >> 8) & 0xFF; } int inline make(int x, int y) { return (x & 0xFF) | ((y & 0xFF) << 8); } bool inline isvalid(int x, int y) { return x >= 0 && y >= 0 && x < N && y < N; } char int2char(int i) { return i < 0 ? ' ' : (i < 10 ? '0' + i : (i < 36 ? 'a' + i - 10 : (i < 62 ? 'A' + i - 36 : '-'))); } void print(int map[N][N]) { for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { cout << int2char(getx(map[i][j])-1) << int2char(gety(map[i][j])); if (j != N - 1) cout << '.'; } cout << endl; } } int main() { static int map[N][N]; static int flg[N][N]; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { map[i][j] = 0; flg[i][j] = 0; } } map[0][2] = -1; map[1][2] = -1; map[2][3] = -1; map[2][4] = -1; map[3][4] = -1; print(map); cout << "---\r\n\r\n" << endl; queue<int> q; flg[0][0] = 1; q.push(make(0, 0)); while (!q.empty()) { int t = q.front(); int x = getx(t); int y = gety(t); int u = 1, d = 1, l = 1, r = 1; for (int step = 1; step < N; step++) { int dist = make(getx(map[x][y])+1, gety(map[x][y])+step); if (u && isvalid(x,y-step) && flg[x][y-step] == 0) { if (map[x][y-step] != -1) { map[x][y-step] = dist; flg[x][y-step] = 1; q.push(make(x,y-step)); } else { u = 0; } } if (d && isvalid(x,y+step) && flg[x][y+step] == 0) { if (map[x][y+step] != -1) { map[x][y+step] = dist; flg[x][y+step] = 1; q.push(make(x,y+step)); } else { d = 0; } } if (l && isvalid(x-step,y) && flg[x-step][y] == 0) { if (map[x-step][y] != -1) { map[x-step][y] = dist; flg[x-step][y] = 1; q.push(make(x-step,y)); } else { l = 0; } } if (r && isvalid(x+step,y) && flg[x+step][y] == 0) { if (map[x+step][y] != -1) { map[x+step][y] = dist; flg[x+step][y] = 1; q.push(make(x+step,y)); } else { r = 0; } } } q.pop(); } print(map); cout << "finish..." << endl; return 0; }