之前随便投了一下今日头条竟然过了,虽然不是很想去,但是已经给了我笔试邀约好歹也就做一下,锻炼锻炼脑子,说句实话还是挺喜欢头条的题目的,做起来很有以前 ACM 刷题的感觉,不像其他公司总感觉"工程化"比较多。

推箱子

第一题比较简单,表面上无从下手但是仔细想想应该能想出来的,本质上跟走迷宫问题本质上是一致的，只不过在这题中状态既要保存人的位置也要保存箱子的位置。暴力搜索能够,AC 代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
char mp[55][55];
bool used[55][55][55][55];
int n, m;
struct Point {
    int x, y;
}you, st, ed;

struct State {
    int x, y, xx, yy, step;
    State() { step = 0; };
    State(int x, int y, int xx, int yy,int step)  {
        this->x = x;
        this->y = y;
        this->xx = xx;
        this->yy = yy;
        this->step = step;
    }
    State(Point a, Point b) {
        x = a.x;
        y = a.y;
        xx = b.x;
        yy = b.y;
        step = 0;
    }
};

int dir[4][2] = { {1,0},{-1,0},{0,1},{0,-1} };
bool valid(int x, int y) {
    if (x<1 || x>n ) return false;
    if (y<1 || y>m ) return false;
    if (mp[x][y] == '#') return false;
    return true;
}

int bfs() {
    queue<State>Q;
    memset(used, 0, sizeof(used));
    State start(you,st);
    Q.push(start);
    used[start.x][start.y][start.xx][start.yy] = 1;
    while (!Q.empty()) {
        State cur = Q.front();
        if (cur.xx == ed.x && cur.yy == ed.y) return cur.step;
        for (int i = 0; i < 4; i++) {
            int x = cur.x + dir[i][0];
            int y = cur.y + dir[i][1];
            int xx = cur.xx;
            int yy = cur.yy;
            if (!valid(x, y))continue;
            if (x == cur.xx && y == cur.yy) {
                xx = cur.xx + dir[i][0];
                yy = cur.yy + dir[i][1];
            }
            if (!valid(xx, yy)) continue;

            if (xx == ed.x && yy == ed.y) {
                return cur.step + 1;
            }
            if (!used[x][y][xx][yy]) {
                used[x][y][xx][yy] = 1;
                Q.push(State(x, y, xx, yy,cur.step + 1));
            }
        }
        Q.pop();
    }
    return -1;
}
int main() {
    while (cin>>n>>m) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                cin >> mp[i][j];
                if (mp[i][j] == 'S') {
                    you.x = i;
                    you.y = j;
                }
                if (mp[i][j] == '0') {
                    st.x = i;
                    st.y = j;
                }
                if (mp[i][j] == 'E') {
                    ed.x = i;
                    ed.y = j;
                }
            }
        }
        cout << bfs() << endl;
    }
}
/*
3 6
.S#..E
.#.0..
......
*/

房间重排

设有 n 个房间，将第 i 个房间的人取出来，依次重新安排到 i+1,i+2,...1,2,3 的房间，不断循环。给你最终各房间人数和最后安排的房间，求最开始的各个房间中人数。

这题有多解，special judge。基本思路就是找到一个包含人数最少的房间，从这个房间开始逆推还原。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
int n, m;
long long a[111111], b[111111];
vector<int>vt;

bool check(int mIndex) {
    for (int i = 0; i < n; i++) {
        if (i == mIndex) b[i] = n * a[i];
        else b[i] = a[i] - a[mIndex];
    }
    for (int i = mIndex; i != m; i = (i + 1) % n) {
        b[i]--;
        b[mIndex]++;
        if (b[i] < 0) return false;
    }
    return true;
}
int main() {
    while (~scanf("%d%d", &n, &m)) {
        long long mVal = -1;
        vt.clear();
        m--;
        for (int i = 0; i < n; i++) {
            cin >> a[i];
            if (mVal == -1 || mVal >= a[i]) {
                if (mVal > a[i])vt.clear();
                mVal = a[i];
                vt.push_back(i);
            }
        }
        for (int i = 0; i < (int)vt.size(); i++) {
            int mIndex = vt[i];
            if (check(mIndex))break;
        }
        for (int i = 0; i < n; i++) {
            printf("%lld%c", b[i], i == n - 1 ? '\n' : ' ');
        }
    }
}
/*
3 1
6 5 1
*/