D. Colored Portals

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52.D. Colored Portals
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收起

https://codeforces.com/contest/2004/problem/D

题意：给定n个字符串，每个字符串有2个字符，如果两个字符串中有相同的字符，则这两个位置互相到达，到达的代价为坐标距离。并且所有字符串的种类只有6种，给定m个查询，求两个地方到达的最小代价。

思路：直接暴力，为6种字符串编号。如果查询位置上的字符串有相同字符，那可以直接算距离。如果没有相同字符，则检查距离l和r最近的其他4种字符串的位置，取代价最小的一个即可。

总结：感觉代码写复杂了

map<string, int> mapp{{"BG", 0}, {"BR", 1}, {"BY", 2}, {"GR", 3}, {"GY", 4}, {"RY", 5}};

void solve(){
    int n, m;
    cin >> n >> m;

    vector<int> a(n);
    vector<string> t(n);
    for (int i = 0; i < n; ++i) {
        string s;
        cin >> s;
        t[i] = s;
        a[i] = mapp[s];
    }
    vector<array<int, 6>> b(n + 1, {-1, -1, -1, -1, -1, -1});
    for (int i = 0; i < n; ++i) {
        b[i + 1] = b[i];
        b[i + 1][a[i]] = i;
    }
    vector<array<int, 6>> c(n + 1, {-1, -1, -1, -1, -1, -1});
    for (int i = n - 2; i >= 0; --i) {
        c[i] = c[i + 1];
        c[i][a[i + 1]] = i + 1;
    }

    while (m --) {
        int l, r;
        cin >> l >> r;
        l --;
        r --;
        if (l > r) {
            swap(l, r);
        }
        if (t[l][0] == t[r][0] || t[l][0] == t[r][1] || t[l][1] == t[r][0] || t[l][1] == t[r][1]) {
            cout << r - l << '\n';
        }
        else {
            int t1 = a[l];
            int t2 = a[r];
            bool ok = false;
            for (int i = 0; i < 6; ++i) {
                if (i != t1 && i != t2) {
                    if (b[l][i] != -1 || b[r][i] != -1 || c[l][i] != -1 || c[r][i] != -1) {
                        ok = true;
                        break;
                    }
                }
            }
            if (!ok) {
                cout << "-1\n";
            }
            else {
                int res = (int)1e9;
                for (int i = 0; i < 6; ++i) {
                    if (i != t1 && i != t2) {
                        if (b[l][i] != -1) {
                            res = min(res, abs(l - b[l][i]) + abs(r - b[l][i]));
                        }
                        if (c[l][i] != -1) {
                            res = min(res, abs(c[l][i] - l) + abs(r - c[l][i]));
                        }
                        if (c[r][i] != -1) {
                            res = min(res, abs(c[r][i] - l) + abs(c[r][i] - r));
                        }
                        if (b[r][i] != -1) {
                            res = min(res, abs(r - b[r][i]) + abs(l - b[r][i]));
                        }
                    }
                }
                cout << res << '\n';
            }
        }
    }
}
posted @   _Yxc  阅读(83)  评论(0)  编辑  收藏  举报
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