POJ 2488 A Knight’s Journey

[★★☆☆☆]搜索 深度优先

• 题目大意：

  给出一个国际棋盘的大小，判断马能否不重复的走过所有格，并记录下其中按字典序排列的第一种路径。

• 样例

  输入：
  3
  1 1
  2 3
  4 3
  输出：
  Scenario #1:
  A1

  Scenario #2:
  impossible

  Scenario #3:
  A1B3C1A2B4C2A3B1C3A4B2C4

• 解题思路：

  水题，根据字典序dfs就行了。

• 代码

#include <iostream>#include <algorithm>using namespace std;int h, l;bool used[30][30];struct point{    int x, y;};point ps[30];int ctp;int dfs(int x, int y) {    if (x-2 >= 1 && y-1 >= 1 && !used[x-2][y-1]) {        used[x-2][y-1] = 1;        point tp = {x-2, y-1};        ps[ctp++] = tp;        if(dfs(x-2, y-1)) return 1;        ctp--;        used[x-2][y-1] = 0;    }    if (x-2 >= 1 && y+1 <= l && !used[x-2][y+1]) {        used[x-2][y+1] = 1;        point tp = {x-2, y+1};        ps[ctp++] = tp;        if(dfs(x-2, y+1)) return 1;        ctp--;        used[x-2][y+1] = 0;    }    if (x-1 >= 1 && y-2 >= 1 && !used[x-1][y-2]) {        used[x-1][y-2] = 1;        point tp = {x-1, y-2};        ps[ctp++] = tp;        if(dfs(x-1, y-2)) return 1;        ctp--;        used[x-1][y-2] = 0;    }    if (x-1 >= 1 && y+2 <= l && !used[x-1][y+2]) {        used[x-1][y+2] = 1;        point tp = {x-1, y+2};        ps[ctp++] = tp;        if(dfs(x-1, y+2)) return 1;        ctp--;        used[x-1][y+2] = 0;    }    if (x+1 <= h && y-2 >= 1 && !used[x+1][y-2]) {        used[x+1][y-2] = 1;        point tp = {x+1, y-2};        ps[ctp++] = tp;        if(dfs(x+1, y-2)) return 1;        ctp--;        used[x+1][y-2] = 0;    }    if (x+1 <= h && y+2 <= l && !used[x+1][y+2]) {        used[x+1][y+2] = 1;        point tp = {x+1, y+2};        ps[ctp++] = tp;        if(dfs(x+1, y+2)) return 1;        ctp--;        used[x+1][y+2] = 0;    }    if (x+2 <= h && y-1 >= 1 && !used[x+2][y-1]) {        used[x+2][y-1] = 1;        point tp = {x+2, y-1};        ps[ctp++] = tp;        if(dfs(x+2, y-1)) return 1;        ctp--;        used[x+2][y-1] = 0;    }    if (x+2 <= h && y+1 <= l && !used[x+2][y+1]) {        used[x+2][y+1] = 1;        point tp = {x+2, y+1};        ps[ctp++] = tp;        if (dfs(x+2, y+1)) return 1;        ctp--;        used[x+2][y+1] = 0;    }    if (ctp == l*h) {        for (int i = 0; i < ctp; i++) {            char c = ps[i].x + 'A' - 1;            int d = ps[i].y;            cout << c << d;        }        cout << endl;        return 1;    }    return 0;}int main() {    int TT;    int ct = 1;    cin >> TT;    while (TT--) {        cin >> l >> h;        ctp = 0;        for (int i = 0; i < 30; i++) {            for (int j = 0; j < 30; j++) {                used[i][j] = 0;            }        }        used[1][1] = 1;        point tp = {1,1};        ps[ctp++] = tp;        cout << "Scenario #" << ct++ << ":" << endl;        if(!dfs(1, 1)) cout << "impossible" << endl;        cout << endl;    }    return 0;}