poj 2488 骑士游历

来源：互联网发布：战斗机飞行员知乎编辑：程序博客网时间：2024/04/29 15:03

深度优先搜索与回溯结合进行

单纯的搜索与回溯的写法是不一样的，

回溯尤其是要记录搜索路径的时候，需要更多的处理

这题目感觉用数组，然后记录搜素深度来记录路径更好，还是用了双端队列，更麻烦了，调了一会

关键：如果DFS能达到，则从任何一点出发都能达到，所以以A1为起点，注意按照字典顺序

#include <iostream>#include <sstream>#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <vector>#include <set>#include <cctype>#include <algorithm>#include <cmath>#include <deque>#include <queue>#include <map>#include <queue>#include <list>#include <iomanip>using namespace std;                                               //////#define INF 0xffffff7#define maxn 30#define max(a,b)(a>b?a:b)///int color[maxn][maxn];//是否访问过int p, q;struct pos{int q;int p;};bool finish;deque<pos> stk;//记录搜索路径int dy[] = {-2, -2, -1, -1,  1, 1,  2, 2};int dx[] = {-1,  1, -2,  2, -2, 2, -1, 1};void dfs(int cur){int i, j;if ((cur == p * q) && !finish){finish = true;while (!stk.empty()){pos tmpPos;tmpPos = stk.front();stk.pop_front();char temp = tmpPos.q + 'A';printf("%c%d", temp, tmpPos.p + 1);}printf("\n");return;}pos curpos = stk.back();for (i = 0; i < 8; i++){pos newpos;newpos.q = curpos.q + dy[i];newpos.p = curpos.p + dx[i];if (newpos.q >= 0 && newpos.q < q && newpos.p >= 0 && newpos.p < p && color[newpos.q][newpos.p] == 0 && !finish){color[newpos.q][newpos.p] = 1;stk.push_back(newpos);dfs(cur + 1);if (!finish)stk.pop_back();color[newpos.q][newpos.p] = 0;}}}int main(){int i, j, k;int ncases;int cases = 1;scanf("%d", &ncases);while (ncases--){scanf("%d %d", &p, &q);//q代表字母个数，p代表数字个数memset(color, 0, sizeof(color));finish = false;while (!stk.empty())stk.pop_back();printf("Scenario #%d:\n", cases);cases++;pos startpos;startpos.q = startpos.p = 0;color[startpos.q][startpos.p] = 1;stk.push_front(startpos);dfs(1);if (!finish)printf("impossible\n");printf("\n");}///    return 0;}