CSP-S 20

9.11

今天是挂分的好日子～～～

101=0+10+91

t1 追逐游戏 (chase)

倍增+k祖先

算了不想写了.......

第2次没保存然后死机丢失了...

不难理解也不难实现，细节手模即可。

t1 就这样吧，看代码。

code

点击查看代码

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, q;
vector<int> e[N];
int fa[N][20], dep[N];

void dfs1(int x, int f)
{
    fa[x][0] = f;
    dep[x] = dep[f] + 1;
    for (int i = 1; i <= 19; ++i)
        fa[x][i] = fa[fa[x][i - 1]][i - 1];
    for (auto y : e[x])
        if (y != f)
            dfs1(y, x);
}

int LCA(int x, int y)
{
    if (dep[x] < dep[y])
        swap(x, y);
    for (int i = 19; ~i; --i)
        if (dep[fa[x][i]] >= dep[y])
            x = fa[x][i];
    if (x == y)
        return x;
    for (int i = 19; ~i; --i)
        if (fa[x][i] != fa[y][i])
            x = fa[x][i], y = fa[y][i];
    return fa[x][0];
}

inline int get_dis(int x, int y)
{
    int lca = LCA(x, y);
    return dep[x] + dep[y] - 2 * dep[lca];
}

inline int jump_k(int x, int k)
{
    for (int i = 19; ~i; --i) //~i --> i>=0!!!!!!!
        if (k & (1 << i))
            x = fa[x][i];
    return x;
}

inline void solve(int x, int y)
{
    int dis = get_dis(x, y);
    int k = (dis + 1) >> 1;
    cout << k << ' ';
    if (dep[x] < dep[y])
        swap(x, y), k = dis - k;
    cout << jump_k(x, k) << "\n";
}

signed main()
{
    freopen("chase.in", "r", stdin);
    freopen("chase.out", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> q;
    for (int i = 1, u, v; i < n; ++i)
    {
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dfs1(1, 0);
    int s1, t, s2;
    while (q--)
    {
        cin >> s1 >> t >> s2;
        int dis1 = get_dis(s1, t);
        int dis2 = get_dis(s2, t);
        if (dis1 < dis2)
            cout << dis2 << ' ' << t << "\n";
        else
            solve(s1, s2);
    }
    return 0;
}

t2 统计

赛时用memset挂了20pts...

正解巧妙。

类似异或哈希做法。

具体的：

对于 \(1\) 到 \(m-1\) 中每个数随机一个哈希值，令 \(v_m=-\sum_{i=1}^{m-1}v_i\)

这样做可使满足要求的区间 \(l,r\) 的 \(sum_r-sum_{l-1}\) 为0.

考虑快速统计答案：

将 \(sum\) 升序排列，之后快速统计。

“快速统计”部分看代码理解（语文功底太差导致的）。

code

点击查看代码

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e6 + 10;
int T;
int n, m;
int a[N], sum[N], v[N];

mt19937_64 myrand(time(0));

signed main()
{
    freopen("st.in", "r", stdin);
    freopen("st.out", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> T;
    while (T--)
    {
        cin >> n >> m;
        for (int i = 0; i <= n; ++i)
            sum[i] = 0, v[i] = 0;
        for (int i = 1; i < m; ++i)
            v[i] = myrand(), v[m] -= v[i];
        for (int i = 1; i <= n; ++i)
            cin >> a[i], sum[i] = sum[i - 1] + v[a[i]];
        sort(sum, sum + 1 + n); // 注意必须手动保证sum[0]为0!!!
        int cnt = 1, ans = 0;
        for (int i = 1; i <= n; ++i)
        {
            if (sum[i] != sum[i - 1])
                cnt = 0;
            ans += cnt++;
        }
        cout << ans << "\n";
    }
    return 0;
}

t3 软件工程

抽象东西假贪心有 96pts!!?

假做法：

code

点击查看代码

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e5 + 10;
const int inf = 1e9 + 7;
int n, k;
bool flag[N];
struct node
{
    int l, r, id;
} a[N], b[N];

inline bool cmp1(node a, node b) { return a.l < b.l; }
inline bool cmp2(node a, node b) { return a.l > b.l; }

signed main()
{
    freopen("se.in", "r", stdin);
    freopen("se.out", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> k;
    for (int i = 1; i <= n; ++i)
    {
        cin >> a[i].l >> a[i].r;
        a[i].id = i;
        b[i].l = a[i].r - a[i].l, b[i].r = i;
    }
    int ans = 0;
    if (k >= n)
    {
        for (int i = 1; i <= n; ++i)
            ans += a[i].r - a[i].l;
        cout << ans << "\n";
        return 0;
    }
    sort(a + 1, a + 1 + n, cmp1);
    sort(b + 1, b + 1 + n, cmp2);
    for (int i = 1; i <= k - 1; ++i)
        ans += b[i].l, flag[b[i].r] = 1;
    int l = 0, r = inf;
    for (int i = 1; i <= n; ++i)
    {
        if (flag[a[i].id])
            continue;
        l = max(l, a[i].l), r = min(r, a[i].r);
    }
    if (l < r)
        ans += r - l;

    memset(flag, 0, sizeof(flag));//下面几乎一样的部分你可以理解成针对hack的特判
    int maxn = 0, dat1 = 0, dat2 = 0;
    // for (int i = 1; i <= n; ++i)
    //     cerr << a[i].l << ' ' << a[i].r << "\n";
    for (int i = 1; i < n; ++i)
    {
        if (min(a[i].r, a[i + 1].r) - a[i + 1].l > maxn)
            maxn = min(a[i].r, a[i + 1].r) - a[i + 1].l, dat1 = a[i].id, dat2 = a[i + 1].id;
    }
    int ans2 = maxn;
    // cerr << dat1 << ' ' << dat2 << "\n";
    flag[dat1] = flag[dat2] = 1;
    for (int i = 1; i <= min(k - 2, n); ++i)
    {
        if (flag[i])
            continue;
        ans2 += b[i].l, flag[b[i].r] = 1;
    }
    l = 0, r = 1e9;
    for (int i = 1; i <= n; ++i)
    {
        if (flag[a[i].id])
            continue;
        l = max(l, a[i].l), r = min(r, a[i].r);
    }
    // cerr << l << ' ' << r << "\n";
    // cerr << ans2 << "\n";
    if (l < r)
        ans2 += r - l;
    cout << max(ans,ans2) << '\n';
    return 0;
}

至于真的：

咕

t4 命运的X

咕

看看题解说的：