pku 1101 bfs

/* //很很很经典的bfs, 这也是我第一次正式用bfs来刷题, 感受颇深: //one.cpp是用bfs来解决从一个点到另一个点的最短路程长度的方法(写的有点冗杂) 注意的几点: 1.为什么用bfs而不是dfs来解决最短路径长度问题? dfs搜索到另一个点后将所经历的点设为visited,不能再次访问; 而bfs是宽度优先的策略,则可以涉及每一条路径. 故,求一个可行解的问题(判断是否有路径)的问题适于用dfs, 最优解,最短路径等求优化解的问题适于用bfs. 2.怎样求最短路径长度? 用一个等大的二维数组记录,记录到达所经过的每一点的路径长度,最先到达的距离最短. //two.cpp 求最短路径上的转弯的线段数 在某一个的方向上所经历的点全部入队(不超界),再让先进的先出... 就可以计算出一共拐了几次弯. (这种方法十分巧妙!) 这类似于游戏"连连看"(一次只能拐两个弯,共三条线段). */ //one.cpp #include <iostream> #include <queue> using namespace std; struct Node { int x,y; }; queue<Node> q; bool map[80][80]; int value[80][80]; int m,n; int x1,y1,x2,y2; int bfs() { Node d; d.x = x1, d.y = y1; q.push(d); value[x1][y1] = 0; while(!q.empty()) { Node temp = q.front(); x1 = temp.x, y1 = temp.y; if(x1 == x2 && y1 == y2) break; q.pop(); if(x1-1 >= 0 && value[x1-1][y1]==-1 && map[x1-1][y1]) { Node d1; d1.x = x1-1, d1.y = y1; q.push(d1); value[x1-1][y1] = value[x1][y1]+1; } if(y1-1 >= 0 && value[x1][y1-1]==-1 && map[x1][y1-1]) { Node d2; d2.x = x1, d2.y = y1-1; q.push(d2); value[x1][y1-1] = value[x1][y1]+1; } if(x1+1 <= m+1 && value[x1+1][y1]==-1 && map[x1+1][y1]) { Node d3; d3.x = x1+1, d3.y = y1; q.push(d3); value[x1+1][y1] = value[x1][y1]+1; } if(y1+1 <= n+1 && value[x1][y1+1]==-1 && map[x1][y1+1]) { Node d4; d4.x = x1, d4.y = y1+1; q.push(d4); value[x1][y1+1] = value[x1][y1]+1; } } return value[x2][y2]; } int main() { int cnt1 = 0; char ch; while(cin >> n >> m && !(n==0&&m==0))//n:列 m:行 { getchar(); for(int j = 0; j <= n+1; ++j) map[0][j] = true; for(int j = 0; j <= n+1; ++j) map[m+1][j] = true; for(int i = 1; i <= m; ++i) { map[i][0] = true; for(int j = 1; j <= n; ++j) { ch = getchar(); if(ch == ' ') map[i][j] = true; else map[i][j] = false; } getchar(); map[i][n+1] = true; } printf("Board #%d:/n", ++cnt1); int cnt2 = 0; while(cin >> y1 >> x1 >> y2 >> x2 && !(x1==0&&y1==0&&x2==0&&y2==0)) { while(!q.empty()) q.pop(); memset(value, -1, sizeof(value)); map[x2][y2] = true; if(bfs() != -1) printf("Pair %d: %d segments./n", ++cnt2, value[x2][y2]-1); else printf("Pair %d: impossible./n", ++cnt2); map[x2][y2] = false; } printf("/n"); } return 0; } //two.cpp #include <iostream> #include <queue> using namespace std; struct Node { int x,y; }; int x[4] = {1, -1, 0, 0}; int y[4] = {0, 0, 1, -1}; queue<Node> q; bool map[80][80]; int value[80][80]; int m,n; int x1,y1,x2,y2; bool ok(int i, int j) { if(i < 0 || j < 0 || i > m+1 || j > n+1) return false; return true; } int bfs() { Node d; d.x = x1, d.y = y1; q.push(d); value[x1][y1] = 0; while(!q.empty() && value[x2][y2] == -1) { for(int i = 0; i < 4; ++i) { int tempx = q.front().x + x[i]; int tempy = q.front().y + y[i]; while(ok(tempx, tempy) && map[tempx][tempy] && value[tempx][tempy] == -1) { Node d; d.x = tempx, d.y = tempy; q.push(d); value[tempx][tempy] = value[q.front().x][q.front().y] + 1; tempx += x[i]; tempy += y[i]; } } q.pop(); } return value[x2][y2]; } int main() { int cnt1 = 0; char ch; while(cin >> n >> m && !(n==0&&m==0))//n:列 m:行 { getchar(); for(int j = 0; j <= n+1; ++j) map[0][j] = true; for(int j = 0; j <= n+1; ++j) map[m+1][j] = true; for(int i = 1; i <= m; ++i) { map[i][0] = true; for(int j = 1; j <= n; ++j) { ch = getchar(); if(ch == ' ') map[i][j] = true; else map[i][j] = false; } getchar(); map[i][n+1] = true; } printf("Board #%d:/n", ++cnt1); int cnt2 = 0; while(cin >> y1 >> x1 >> y2 >> x2 && !(x1==0&&y1==0&&x2==0&&y2==0)) { while(!q.empty()) q.pop(); memset(value, -1, sizeof(value)); map[x2][y2] = true; if(bfs() != -1) printf("Pair %d: %d segments./n", ++cnt2, value[x2][y2]); else printf("Pair %d: impossible./n", ++cnt2); map[x2][y2] = false; } printf("/n"); } return 0; }