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）：

 #include <bits/stdc++.h>
 using namespace std;

 const int MAXV = ;

 const char str_op[] = "dulr";
 int fx[] = {, -, , };
 int fy[] = {, , -, };

 inline int id(char c) {
     static const char op[] = "dulr";
     return strchr(op, c) - op;
 }

 char op[MAXV];
 int len;
 char mat[][];
 int n, m, s;

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

 void init() {
     for(int i = ; i <= n; ++i)
         for(int j = ; j <= m; ++j) {
             if(mat[i][j] == '') sx = i, sy = j;
             if(mat[i][j] == '') ex = i, ey = j;
             if(mat[i][j] == '') vx = i, vy = j;
         }
 }

 bool solve() {
     int x1 = sx, y1 = sy, x2 = vx, y2 = vy;
     for(int i = ; i < len; ++i) {
         int f = id(op[i]);
         int new_x = x1 + fx[f], new_y = y1 + fy[f];
         if(new_x == x2 && new_y == y2) {
             int pp = x2 + fx[f], qq = y2 + fy[f];
             if(mat[pp][qq] != '') {
                 x1 = new_x, y1 = new_y;
                 x2 = pp,    y2 = qq;
             }
         } else if(mat[new_x][new_y] != '') {
             x1 = new_x, y1 = new_y;
         }
         if(x2 == ex && y2 == ey) return true;
     }
     return false;
 }

 int main() {
     memset(mat, '', sizeof(mat));
     scanf("%d%d%d", &m, &n, &s);
     for(int i = ; i <= n; ++i)
         for(int j = ; j <= m; ++j) scanf(" %c", &mat[i][j]);
     init();
     while(s--) {
         scanf("%d %s", &len, op);
         puts(solve() ? "YES" : "NO");
     }
 }

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）：

 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <iostream>
 using namespace std;
 typedef long long LL;

 int f[][] = {
 {, , },
 {, , },
 {, , },
 {, , },
 {, , },
 {, , },
 {, , },
 {, , }
 };

 char mat[];
 int wino[], winx[];
 int T;

 void build_check(int win[], char c) {
     for(int i = ; i < ; ++i) {
         win[i] = ;
         for(int j = ; j < ; ++j)
             if(mat[f[i][j]] != c) win[i] = ;
     }
 }

 int build_count(char c) {
     return count(mat, mat + , c);
 }

 bool next_win(char c) {
     int win[];
     for(int i = ; i < ; ++i) if(mat[i] == '_') {
         mat[i] = c;
         build_check(win, c);
         if(count(win, win + , )) return true;
         mat[i] = '_';
     }
     return false;
 }

 int solve() {
     int xcnt = build_count('X'), ocnt = build_count('O');
     if(xcnt != ocnt && xcnt -  != ocnt) return puts("Invalid");

     build_check(winx, 'X');
     build_check(wino, 'O');
     if(count(winx, winx + , ) >  && count(wino, wino + , ) > ) return puts("Invalid");
     if(count(winx, winx + , ) >  && xcnt == ocnt) return puts("Invalid");
     if(count(wino, wino + , ) >  && xcnt -  == ocnt) return puts("Invalid");

     if(count(winx, winx + , ) > ) return puts("X win");
     if(count(wino, wino + , ) > ) return puts("O win");

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

     if(next_win(xcnt == ocnt ? 'X' : 'O')) return puts("Next win");
     return puts("Next cannot win");
 }

 int main() {
     scanf("%d", &T);
     while(T--) {
         scanf("%s", mat);
         scanf("%s", mat + );
         scanf("%s", mat + );
         solve();
     }
 }

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）：

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <queue>
 using namespace std;

 const int MAXN = ;
 const int INF = 0x3f3f3f3f;

 struct Node {
     int f, x, y;
     Node() {}
     Node(int f, int x, int y):
         f(f), x(x), y(y) {}
 };

 int fx[] = {-, , , };
 int fy[] = {, , , -};

 int dis[][MAXN][MAXN];
 int mat[MAXN][MAXN];
 bool vis[MAXN][MAXN];
 int T, n, m;

 int bfs(int x1, int y1, int k) {
     memset(vis, , sizeof(vis)); vis[x1][y1] = true;
     memset(dis, 0x3f, sizeof(dis));
     queue<Node> *now = new queue<Node>();
     queue<Node> *nxt = new queue<Node>();
     int step = , res = ;
     for(int i = ; i < ; ++i) now->push(Node(i, x1, y1));
     for(int i = ; i < ; ++i) dis[i][x1][y1] = ;
     while(step <= k && !now->empty()) {
         Node t = now->front(); now->pop();
         if(dis[t.f][t.x][t.y] != step) continue;
         for(int i = ; i < ; ++i) {
             if((t.f + ) %  == i) continue;
             int x = t.x + fx[i], y = t.y + fy[i], d = dis[t.f][t.x][t.y] + (t.f != i);
             if(mat[x][y] == mat[x1][y1] && !vis[x][y] && d <= k) {
                 vis[x][y] = true;
                 res++;
             }
             if(mat[x][y] ==  && d < dis[i][x][y]) {
                 dis[i][x][y] = d;
                 if(t.f == i) now->push(Node(i, x, y));
                 else nxt->push(Node(i, x, y));
             }
         }
         if(now->empty()) {
             step++;
             swap(now, nxt);
         }
     }
     return res;
 }

 int main() {
     scanf("%d", &T);
     while(T--) {
         scanf("%d%d", &n, &m);
         memset(mat, 0x3f, sizeof(mat));
         for(int i = ; i <= n + ; ++i)
             for(int j = ; j <= m + ; ++j) scanf("%d", &mat[i][j]);
         for(int i = ; i <= n + ; ++i)
             for(int j = ; j <= m + ; ++j)
                 if(mat[i][j] == INF) mat[i][j] = ;

         int x, y, k;
         scanf("%d%d%d", &x, &y, &k);
         printf("%d\n", bfs(x + , y + , k));
     }
 }