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

 

posted @ 2015-04-08 22:35  Oyking  阅读(4675)  评论(4编辑  收藏