hihocoder 网易游戏2016实习生招聘在线笔试 解题报告

比赛笔试链接：http://hihocoder.com/contest/ntest2015april/problems

 

题目就不贴了。

 

1、推箱子。

思路：纯模拟。

代码（28MS）：

 1 #include <bits/stdc++.h>

 2 using namespace std;

 3 

 4 const int MAXV = 10010;

 5 

 6 const char str_op[] = "dulr";

 7 int fx[] = {1, -1, 0, 0};

 8 int fy[] = {0, 0, -1, 1};

 9 

10 inline int id(char c) {

11     static const char op[] = "dulr";

12     return strchr(op, c) - op;

13 }

14 

15 char op[MAXV];

16 int len;

17 char mat[105][105];

18 int n, m, s;

19 

20 int sx, sy, vx, vy, ex, ey;

21 

22 void init() {

23     for(int i = 1; i <= n; ++i)

24         for(int j = 1; j <= m; ++j) {

25             if(mat[i][j] == '1') sx = i, sy = j;

26             if(mat[i][j] == '2') ex = i, ey = j;

27             if(mat[i][j] == '3') vx = i, vy = j;

28         }

29 }

30 

31 bool solve() {

32     int x1 = sx, y1 = sy, x2 = vx, y2 = vy;

33     for(int i = 0; i < len; ++i) {

34         int f = id(op[i]);

35         int new_x = x1 + fx[f], new_y = y1 + fy[f];

36         if(new_x == x2 && new_y == y2) {

37             int pp = x2 + fx[f], qq = y2 + fy[f];

38             if(mat[pp][qq] != '4') {

39                 x1 = new_x, y1 = new_y;

40                 x2 = pp,    y2 = qq;

41             }

42         } else if(mat[new_x][new_y] != '4') {

43             x1 = new_x, y1 = new_y;

44         }

45         if(x2 == ex && y2 == ey) return true;

46     }

47     return false;

48 }

49 

50 int main() {

51     memset(mat, '4', sizeof(mat));

52     scanf("%d%d%d", &m, &n, &s);

53     for(int i = 1; i <= n; ++i)

54         for(int j = 1; j <= m; ++j) scanf(" %c", &mat[i][j]);

55     init();

56     while(s--) {

57         scanf("%d %s", &len, op);

58         puts(solve() ? "YES" : "NO");

59     }

60 }

View Code

 

2、井字棋

思路：俗称井字过三关。题目没有提到的三种不合法情况：

①XO都有3连

②X有3连，但是count(X)=count(O)

③O有3连，但是count(X)-1=count(O) 

其他不难。人工手动枚举大法好。简单粗暴不易出错。

不过我又成功在通过样例之前把代码交了上去导致WA20（不算罚时就是好啊）

代码（15MS）：

 1 #include <cstdio>

 2 #include <algorithm>

 3 #include <cstring>

 4 #include <iostream>

 5 using namespace std;

 6 typedef long long LL;

 7 

 8 int f[8][3] = {

 9 {0, 1, 2},

10 {3, 4, 5},

11 {6, 7, 8},

12 {0, 3, 6},

13 {1, 4, 7},

14 {2, 5, 8},

15 {0, 4, 8},

16 {2, 4, 6}

17 };

18 

19 char mat[10];

20 int wino[8], winx[8];

21 int T;

22 

23 void build_check(int win[], char c) {

24     for(int i = 0; i < 8; ++i) {

25         win[i] = 1;

26         for(int j = 0; j < 3; ++j)

27             if(mat[f[i][j]] != c) win[i] = 0;

28     }

29 }

30 

31 int build_count(char c) {

32     return count(mat, mat + 9, c);

33 }

34 

35 bool next_win(char c) {

36     int win[8];

37     for(int i = 0; i < 9; ++i) if(mat[i] == '_') {

38         mat[i] = c;

39         build_check(win, c);

40         if(count(win, win + 8, 1)) return true;

41         mat[i] = '_';

42     }

43     return false;

44 }

45 

46 int solve() {

47     int xcnt = build_count('X'), ocnt = build_count('O');

48     if(xcnt != ocnt && xcnt - 1 != ocnt) return puts("Invalid");

49 

50     build_check(winx, 'X');

51     build_check(wino, 'O');

52     if(count(winx, winx + 8, 1) > 0 && count(wino, wino + 8, 1) > 0) return puts("Invalid");

53     if(count(winx, winx + 8, 1) > 0 && xcnt == ocnt) return puts("Invalid");

54     if(count(wino, wino + 8, 1) > 0 && xcnt - 1 == ocnt) return puts("Invalid");

55 

56     if(count(winx, winx + 8, 1) > 0) return puts("X win");

57     if(count(wino, wino + 8, 1) > 0) return puts("O win");

58 

59     if(build_count('_') == 0) return puts("Draw");

60 

61     if(next_win(xcnt == ocnt ? 'X' : 'O')) return puts("Next win");

62     return puts("Next cannot win");

63 }

64 

65 int main() {

66     scanf("%d", &T);

67     while(T--) {

68         scanf("%s", mat);

69         scanf("%s", mat + 3);

70         scanf("%s", mat + 6);

71         solve();

72     }

73 }

View Code

 

3、连连看

思路：参照最小转弯问题：http://www.cnblogs.com/oyking/p/3756208.html

估计直接来个heap+dijkstra也能100分把。

代码（302MS）：

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <iostream>

 4 #include <algorithm>

 5 #include <queue>

 6 using namespace std;

 7 

 8 const int MAXN = 210;

 9 const int INF = 0x3f3f3f3f;

10 

11 struct Node {

12     int f, x, y;

13     Node() {}

14     Node(int f, int x, int y):

15         f(f), x(x), y(y) {}

16 };

17 

18 int fx[] = {-1, 0, 1, 0};

19 int fy[] = {0, 1, 0, -1};

20 

21 int dis[4][MAXN][MAXN];

22 int mat[MAXN][MAXN];

23 bool vis[MAXN][MAXN];

24 int T, n, m;

25 

26 int bfs(int x1, int y1, int k) {

27     memset(vis, 0, sizeof(vis)); vis[x1][y1] = true;

28     memset(dis, 0x3f, sizeof(dis));

29     queue<Node> *now = new queue<Node>();

30     queue<Node> *nxt = new queue<Node>();

31     int step = 0, res = 0;

32     for(int i = 0; i < 4; ++i) now->push(Node(i, x1, y1));

33     for(int i = 0; i < 4; ++i) dis[i][x1][y1] = 0;

34     while(step <= k && !now->empty()) {

35         Node t = now->front(); now->pop();

36         if(dis[t.f][t.x][t.y] != step) continue;

37         for(int i = 0; i < 4; ++i) {

38             if((t.f + 2) % 4 == i) continue;

39             int x = t.x + fx[i], y = t.y + fy[i], d = dis[t.f][t.x][t.y] + (t.f != i);

40             if(mat[x][y] == mat[x1][y1] && !vis[x][y] && d <= k) {

41                 vis[x][y] = true;

42                 res++;

43             }

44             if(mat[x][y] == 0 && d < dis[i][x][y]) {

45                 dis[i][x][y] = d;

46                 if(t.f == i) now->push(Node(i, x, y));

47                 else nxt->push(Node(i, x, y));

48             }

49         }

50         if(now->empty()) {

51             step++;

52             swap(now, nxt);

53         }

54     }

55     return res;

56 }

57 

58 int main() {

59     scanf("%d", &T);

60     while(T--) {

61         scanf("%d%d", &n, &m);

62         memset(mat, 0x3f, sizeof(mat));

63         for(int i = 2; i <= n + 1; ++i)

64             for(int j = 2; j <= m + 1; ++j) scanf("%d", &mat[i][j]);

65         for(int i = 1; i <= n + 2; ++i)

66             for(int j = 1; j <= m + 2; ++j)

67                 if(mat[i][j] == INF) mat[i][j] = 0;

68 

69         int x, y, k;

70         scanf("%d%d%d", &x, &y, &k);

71         printf("%d\n", bfs(x + 1, y + 1, k));

72     }

73 }

View Code