2022河南萌新联赛第（四）场：郑州轻工业大学_2022河南萌新第四场f-优快云博客

本文链接：https://blog.youkuaiyun.com/god_libl/article/details/126468977

本文探讨了如何利用并查集数据结构解决字符串中的连通块问题，通过观察特定字符（Z, U, L, I）进行分批合并，优化了时间复杂度至O(4*n)，同时展示了大盗问题的bitset优化和密集数据的二分查找解决方案。

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A-并查集

// 考察连通块，一眼想到并查集
// 对于并查集合 (a, b)合并之后合并(b, c) 与 (a, b)合并之后合并(a, c) 是没有区别的
// 所以拿第一个 Z 向后合并一遍，遇见 Z 、U 、 L、 I 就合并；
// 再拿第一个 U 按照规则向后合并一遍，再拿第一个 L 按照规则向后合并一遍，再拿第一个 I 按照规则向后合并一遍即可
// 时间复杂度为O(4 * n)
const int N = 1e6 + 10;
char str[N];
int p[N], s[N];

int find(int x)
{
    if(x != p[x]) p[x] = find(p[x]);
    return p[x];
}

int main()
{
    IOS;
#ifdef LIBOLIN
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    cin >> (str + 1);
    
    int n = strlen(str + 1);
    int a, b, c, d;
    a = b = c = d = n + 1;
    for (int i = 1; i <= n; i++)
        if (str[i] == 'Z')
        {
            a = i;
            break;
        }
    for (int i = 1; i <= n; i++)
        if (str[i] == 'U')
        {
            b = i;
            break;
        }
    for (int i = 1; i <= n; i++)
        if (str[i] == 'L')
        {
            c = i;
            break;
        }
    for (int i = 1; i <= n; i++)
        if (str[i] == 'I')
        {
            d = i;
            break;
        }

    for(int i = 1; i <= n; i ++ )
        p[i] = i, s[i] = 1;

    for(int i = a + 1; i <= n; i ++ )
    {
        if(str[i] == 'Z' || str[i] == 'U' || str[i] == 'L' || str[i] == 'I')
        {
            int pa = find(a);
            int pb = find(i);
            if(pa != pb)
            {
                s[pb] += s[pa];
                p[pa] = pb;
            }
        }
    }
    for(int i = b + 1; i <= n; i ++ )
    {
        if(str[i] == 'U' || str[i] == 'L' || str[i] == 'I')
        {
            int pa = find(b);
            int pb = find(i);
            if(pa != pb)
            {
                s[pb] += s[pa];
                p[pa] = pb;
            }
        }
    }
    for(int i = c + 1; i <= n; i ++ )
    {
        if(str[i] == 'L' || str[i] == 'I')
        {
            int pa = find(c);
            int pb = find(i);
            if(pa != pb)
            {
                s[pb] += s[pa];
                p[pa] = pb;
            }
        }
    }
    for(int i = d + 1; i <= n; i ++ )
    {
        if(str[i] == 'I')
        {
            int pa = find(d);
            int pb = find(i);
            if(pa != pb)
            {
                s[pb] += s[pa];
                p[pa] = pb;
            }
        }
    }

    int ans = 0;
    for(int i = 1; i <= n; i ++ )
    {
        ans = max(ans, s[i]);
    }
    cout << ans << endl;
    return 0;
}

D-大盗

// 使用bitset容器，优化时间复杂度为O(n*k / 32)

bitset<50010> b = 1;

int main()
{
    IOS;
#ifdef LIBOLIN
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < n; i++)
    {
        int o, w;
        cin >> o >> w;
        if (o == 1)
        {
            b |= (b << w);
        }
        else
        {
            if (b[w])
            {
                b &= 1;
                b[w] = 1;
            }
            else
            {
                b &= 1;
            }
        }
    }
    for (int i = k; i >= 0; i--)
    {
        if (b[i])
        {
            cout << i << endl;
            break;
        }
    }
    return 0;
}

I-密集

// 总结：具有单调性质的数据总可以使用二分
// 坑点：二分答案的时候不要把r开太大要不然会爆ll

const int N = 1e5 + 10;
ll n, k;

struct D
{
    ll s;
    ll v;
    bool operator<(const struct D &W) const
    {
        return s < W.s;
    }
} d[N];

bool check(ll t)
{
    ll maxx = 0, ans = 0;
    for (int i = 0; i < n; i++)
    {
        ll dis = d[i].s + d[i].v * t;
        if(maxx > dis) ans ++;
        maxx = max(maxx, dis);
    }
    return n - ans <= k;
}

int main()
{
    IOS;
#ifdef LIBOLIN
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    cin >> n >> k;
    for (int i = 0; i < n; i++)
        cin >> d[i].s;
    for (int i = 0; i < n; i++)
        cin >> d[i].v;   
    sort(d, d + n);
    ll l = 0, r = 1e9 + 10;
    while (l < r)
    {
        ll mid = (l + r) / 2;
        if (check(mid))
            r = mid;
        else
            l = mid + 1;
    }

    if (l == 1e9 + 10)
        cout << "Never!" << endl;
    else
        cout << l << endl;
    return 0;
}