Problem Description
有一个m*n格的迷宫(表示有m行、n列),其中有可走的也有不可走的,如果用1表示可以走,0表示不可以走,输入这m*n个数据和起始点、结束点(起始点和结束点都是用两个数据来描述的,分别表示这个点的行号和列号)。现在要你编程找出所有可行的道路,要求所走的路中没有重复的点,走时只能是上下左右四个方向。如果一条路都不可行,则输出相应信息(用-1表示无路)。
Input
第一行是两个数m,n(1< m, n< 15),接下来是m行n列由1和0组成的数据,最后两行是起始点和结束点。
Output
所有可行的路径,输出时按照左上右下的顺序。描述一个点时用(x,y)的形式,除开始点外,其他的都要用“->”表示。如果没有一条可行的路则输出-1。
Sample Input
5 4 1 1 0 0 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 5 4
Sample Output
(1,1)->(1,2)->(2,2)->(2,3)->(3,3)->(3,2)->(4,2)->(4,1)->(5,1)->(5,2)->(5,3)->(5,4) (1,1)->(1,2)->(2,2)->(2,3)->(3,3)->(3,2)->(4,2)->(5,2)->(5,3)->(5,4) (1,1)->(1,2)->(2,2)->(3,2)->(4,2)->(4,1)->(5,1)->(5,2)->(5,3)->(5,4) (1,1)->(1,2)->(2,2)->(3,2)->(4,2)->(5,2)->(5,3)->(5,4) (1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,2)->(4,2)->(4,1)->(5,1)->(5,2)->(5,3)->(5,4) (1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,2)->(4,2)->(5,2)->(5,3)->(5,4) (1,1)->(2,1)->(2,2)->(3,2)->(4,2)->(4,1)->(5,1)->(5,2)->(5,3)->(5,4) (1,1)->(2,1)->(2,2)->(3,2)->(4,2)->(5,2)->(5,3)->(5,4)
Hint
#include<stdio.h> #include<string.h> struct node { int x, y; }ls[1000]; int dx[] = {0,-1,0,1}, dy[] = {-1,0,1,0}; int bj[16][16], map[16][16]; int m, n, step, sum = 0, xz, yz; void dfs(int x1, int y1) { int i; if(x1 == xz && y1 == yz) { sum++; for(i = 0; i < step; i++) { printf("(%d,%d)",ls[i].x,ls[i].y); if(i < step - 1) printf("->"); } printf("\n"); } else { int kx, ky; for(i = 0; i < 4; i++) { kx = x1 + dx[i]; ky = y1 + dy[i]; if(kx >= 1 && ky >= 1 && kx <= n && ky <= m && !bj[kx][ky] && map[kx][ky]) { ls[step].x = kx; ls[step].y = ky; step++; bj[kx][ky] = 1; dfs(kx,ky); bj[kx][ky] = 0; step--; } } } } int main() { int i, j,xq,yq; while(~scanf("%d%d",&n,&m)) { memset(bj,0,sizeof(bj)); memset(ls,0,sizeof(ls)); for(i = 1; i <= n; i++) for(j = 1; j <= m; j++) scanf("%d",&map[i][j]); scanf("%d%d%d%d",&xq,&yq,&xz,&yz); ls[0].x = xq; ls[0].y = yq; sum = 0; step = 1; bj[xq][yq] = 1; dfs(xq,yq); if(sum == 0) printf("-1\n"); } return 0; }