走迷宫问题以及C++程序代码.doc

走迷宫M*n格的迷宫，1表示可以走，0表示不可以走读入：m*n个数据和起始点、结束点。先要求找出所有克星的道路，要求所有的路中没有重复的店，走失只能是上下左右四个方向。如果一条路都不行，则输出-1。第一行是两个数m和n(m,n”表示连接没有路径输出-1样例输入：5 41 1 0 01 0 0 00 1 1 01 1 0 11 1 1 11 15 4样例输出：-1样例输入2：5 41 1 0 01 1 1 10 1 1 01 1 0 11 1 1 11 15 4样例输出2：(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)#includeusing namespace std;int main()char names2050;int count=0;for(;)char temp50;cintemp;if(strcmp(temp,#)=0) break; int i;for(i=0;icount;i+)int res=strcmp(temp,namesi);if(resi;j-)strcpy(namesj,namesj-1);strcpy(namesi,temp);count+;break;else if (res=0)break;if(i=count)strcpy(namescount,temp);count+;for(int i=0;icount;i+)coutnamesiendl;return 0;