[ICPC2020南京E] Evil Coordinate

[ICPC2020南京E] Evil Coordinate - 结论

Description

机器人从（0，0）出发，根据题目的输入字符串，到达终点，在已知的路径中会有一个地雷，要求在不改变字符串数量的前提下改变顺序，不踩到地雷。

Solution

只有一个地雷，我们将上下左右各自连在一起，那么4个方向的全排列对应最多4种解，一定包含了所有有解的情况

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

#define int long long

const int dx[5] = {0, -1, 1, 0, 0};
const int dy[5] = {0, 0, 0, 1, -1};
const char lut[5] = {' ', 'L', 'R', 'U', 'D'};

bool check(const vector<int> &seq, int mx, int my)
{
    int x = 0, y = 0;
    if (x == mx && y == my)
        return false;
    for (int i = 0; i < seq.size(); i++)
    {
        x += dx[seq[i]];
        y += dy[seq[i]];
        if (x == mx && y == my)
            return false;
    }
    return true;
}

void solve()
{
    int mx, my;
    cin >> mx >> my;
    string str;
    cin >> str;
    int n = str.length();
    vector<int> seq(n);
    for (int i = 0; i < n; i++)
    {
        if (str[i] == 'L')
            seq[i] = 1;
        if (str[i] == 'R')
            seq[i] = 2;
        if (str[i] == 'U')
            seq[i] = 3;
        if (str[i] == 'D')
            seq[i] = 4;
    }
    vector<int> cnt(6);
    for (int i = 0; i < n; i++)
        cnt[seq[i]]++;
    int a[4];
    iota(a, a + 4, 1);
    do
    {
        vector<int> tseq;
        for (int i = 0; i < 4; i++)
        {
            int c = a[i];
            for (int j = 0; j < cnt[c]; j++)
                tseq.push_back(c);
        }
        // cout << "test ";
        // for (int i = 0; i < n; i++)
        //     cout << lut[tseq[i]];
        // cout << endl;
        if (check(tseq, mx, my))
        {
            for (int i = 0; i < n; i++)
                cout << lut[tseq[i]];
            cout << endl;
            return;
        }
    } while (next_permutation(a, a + 4));
    cout << "Impossible" << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--)
        solve();
}