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 = 10010;

const char str_op[] = "dulr";
int fx[] = {1, -1, 0, 0};
int fy[] = {0, 0, -1, 1};

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

char op[MAXV];
int len;
char mat[105][105];
int n, m, s;

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

void init() {
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) {
            if(mat[i][j] == '1') sx = i, sy = j;
            if(mat[i][j] == '2') ex = i, ey = j;
            if(mat[i][j] == '3') vx = i, vy = j;
        }
}

bool solve() {
    int x1 = sx, y1 = sy, x2 = vx, y2 = vy;
    for(int i = 0; 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] != '4') {
                x1 = new_x, y1 = new_y;
                x2 = pp,    y2 = qq;
            }
        } else if(mat[new_x][new_y] != '4') {
            x1 = new_x, y1 = new_y;
        }
        if(x2 == ex && y2 == ey) return true;
    }
    return false;
}

int main() {
    memset(mat, '4', sizeof(mat));
    scanf("%d%d%d", &m, &n, &s);
    for(int i = 1; i <= n; ++i)
        for(int j = 1; 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[8][3] = {
{0, 1, 2},
{3, 4, 5},
{6, 7, 8},
{0, 3, 6},
{1, 4, 7},
{2, 5, 8},
{0, 4, 8},
{2, 4, 6}
};

char mat[10];
int wino[8], winx[8];
int T;

void build_check(int win[], char c) {
    for(int i = 0; i < 8; ++i) {
        win[i] = 1;
        for(int j = 0; j < 3; ++j)
            if(mat[f[i][j]] != c) win[i] = 0;
    }
}

int build_count(char c) {
    return count(mat, mat + 9, c);
}

bool next_win(char c) {
    int win[8];
    for(int i = 0; i < 9; ++i) if(mat[i] == '_') {
        mat[i] = c;
        build_check(win, c);
        if(count(win, win + 8, 1)) return true;
        mat[i] = '_';
    }
    return false;
}

int solve() {
    int xcnt = build_count('X'), ocnt = build_count('O');
    if(xcnt != ocnt && xcnt - 1 != ocnt) return puts("Invalid");

    build_check(winx, 'X');
    build_check(wino, 'O');
    if(count(winx, winx + 8, 1) > 0 && count(wino, wino + 8, 1) > 0) return puts("Invalid");
    if(count(winx, winx + 8, 1) > 0 && xcnt == ocnt) return puts("Invalid");
    if(count(wino, wino + 8, 1) > 0 && xcnt - 1 == ocnt) return puts("Invalid");

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

    if(build_count('_') == 0) 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 + 3);
        scanf("%s", mat + 6);
        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 = 210;
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[] = {-1, 0, 1, 0};
int fy[] = {0, 1, 0, -1};

int dis[4][MAXN][MAXN];
int mat[MAXN][MAXN];
bool vis[MAXN][MAXN];
int T, n, m;

int bfs(int x1, int y1, int k) {
    memset(vis, 0, 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 = 0, res = 0;
    for(int i = 0; i < 4; ++i) now->push(Node(i, x1, y1));
    for(int i = 0; i < 4; ++i) dis[i][x1][y1] = 0;
    while(step <= k && !now->empty()) {
        Node t = now->front(); now->pop();
        if(dis[t.f][t.x][t.y] != step) continue;
        for(int i = 0; i < 4; ++i) {
            if((t.f + 2) % 4 == 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] == 0 && 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 = 2; i <= n + 1; ++i)
            for(int j = 2; j <= m + 1; ++j) scanf("%d", &mat[i][j]);
        for(int i = 1; i <= n + 2; ++i)
            for(int j = 1; j <= m + 2; ++j)
                if(mat[i][j] == INF) mat[i][j] = 0;

        int x, y, k;
        scanf("%d%d%d", &x, &y, &k);
        printf("%d
", bfs(x + 1, y + 1, k));
    }
}