A. 試卷

B. 果實

據說是另一道題的弱化版，原題帶修改，好像需要用 set 維護節點。這題數據非常友善，可以用莫隊水過（塊長卡了 \(n^{\frac{1}{2}}\) 但沒卡 \(n^{\frac{2}{3}}\) ）。
問題可以用 dfn 序轉化為求區間不同元素個數。離線做。將詢問根據左端點排序，然後維護一個 BIT ，在某個元素第一次出現的位置處 +1 。然後在getsum 就可以得到答案了。左端點右移時，把 BIT 上當前位置 -1 ，在當前位置元素下一次出現的位置（用類似鏈式前向星的方法預處理） +1 即可。
然而考試結束後 OBlack 說在線更好做。樹上差分即可。比如某元素的出現的 dfn 值分別為 \(1,3,10,20\)

離線 BIT 做法

#include <cstdio>
#include <algorithm>
#include <ctype.h>
#include <vector>

using namespace std;

char *p1, *p2, buf[1 << 20], sss[50];

inline char gc()
{
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin))==p1?EOF:*p1++;
}

template<typename T>
void rd(T &num)
{
    char tt;
    while (!isdigit(tt = gc()));
    num = tt - '0';
    while (isdigit(tt = gc()))
        num = num * 10 + tt - '0';
    return;
}

template<typename T>
void pt(T num)
{
    int top = 0;
    do sss[++top] = num % 10 + '0';
    while (num /= 10);
    while (top)
        putchar(sss[top--]);
    putchar('\n');
    return;
}

const int _N = 501000;

vector<int> G[_N];
int dfn[_N], siz[_N], A[_N], B[_N], bit[_N], nxt[_N], fst[_N], ans[_N];
int Time, N, M, C;

struct data {
    int id, v;
    bool operator < (const data &tmp)
    const
    {
        return dfn[v] < dfn[tmp.v];
    }
} Q[_N];

void build(int p, int dad)
{
    dfn[p] = ++Time, siz[p] = 1, B[dfn[p]] = A[p];
    for (int i = G[p].size() - 1; i >= 0; --i) {
        int v = G[p][i];
        if (v != dad)
            build(v, p), siz[p] += siz[v];
    }
    return;
}

void add(int k, int d)
{
    for (int i = k; i <= N; i += i & -i)
        bit[i] += d;
    return;
}

int getsum(int k)
{
    int sum = 0;
    for (int i = k; i; i ^= i & -i)
        sum += bit[i];
    return sum;
}

int main()
{
    
    rd(N), rd(M), rd(C);
    for (int i = 1; i <= N; ++i)
        rd(A[i]);
    for (int a, b, i = 1; i < N; ++i) {
        rd(a), rd(b);
        G[a].push_back(b), G[b].push_back(a);
    }
    build(1, 0);
    for (int i = N; i >= 1; --i)
        nxt[i] = fst[B[i]], fst[B[i]] = i;
    for (int i = 1; i <= N; ++i)
        if (fst[B[i]] == i)
            add(i, 1);
    for (int t, i = 1; i <= M; ++i)
        rd(Q[i].v), Q[i].id = i;
    sort(Q + 1, Q + 1 + M);
    int pos = 1;
    for (int i = 1; i <= M; ++i) {
        while (pos < dfn[Q[i].v]) {
            add(pos, -1);
            if (nxt[pos]) add(nxt[pos], 1);
            ++pos;
        }
        ans[Q[i].id] = getsum(dfn[Q[i].v] + siz[Q[i].v] - 1);
    }
    for (int i = 1; i <= M; ++i)
        pt(ans[i]);
    return 0;
}
 

C. 旅行

#include <cstdio>
#include <algorithm>
#include <ctype.h>
#include <vector>
#include <queue>

using namespace std;

char *p1, *p2, buf[1 << 20], sss[50];

inline char gc()
{
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin))==p1?EOF:*p1++;
}

template<typename T>
void rd(T &num)
{
    char tt;
    while (!isdigit(tt = gc()));
    num = tt - '0';
    while (isdigit(tt = gc()))
        num = num * 10 + tt - '0';
    return;
}

template<typename T>
void pt(T num)
{
    int top = 0;
    do sss[++top] = num % 10 + '0';
    while (num /= 10);
    while (top)
        putchar(sss[top--]);
    putchar('\n');
    return;
}

const int _N = 51000;

struct edge {
    int v, w;
    edge(int v = 0, int w = 0):
        v(v), w(w) { }
    bool operator < (const edge &tmp)
    const
    {
        return w > tmp.w;
    }
};

priority_queue<edge> Q;
vector<edge> G[_N];
int dis[_N];
int N, M, ans = -1;

int dijkstra(int lim, int t)
{
    for (int i = 0; i <= N; ++i)
        dis[i] = -1;
    dis[1] = 0, Q.push(edge(1, 0));
    while (!Q.empty()) {
        edge p = Q.top();
        Q.pop();
        if (p.w != dis[p.v]) continue;
        for (int i = G[p.v].size() - 1; i >= 0; --i) {
            edge v = G[p.v][i];
            
            if (p.v == 1 && v.v >> lim & 1 ^ t) continue;
            if (v.v == 0 && p.v >> lim & 1 ^ t ^ 1) continue;
            
            if (dis[v.v] == -1 || dis[v.v] > dis[p.v] + v.w)
                Q.push(edge(v.v, dis[v.v] = dis[p.v] + v.w));
        }
    }
    return dis[0];
}

int main()
{
    rd(N), rd(M);
    for (int x, y, c, d, i = 1; i <= M; ++i) {
        rd(x), rd(y), rd(c), rd(d);
        if (x > y) swap(x, y), swap(c, d);
        G[x].push_back(edge(y, c));
        if (x == 1) x = 0;
        G[y].push_back(edge(x, d));
    }
    int tmp = ,N cnt = 0;
    while (tmp) ++cnt, tmp >>= 1;
    for (int i = 0; i < cnt; ++i) {
        int t = dijkstra(i, 0);
        if (t != -1 && (ans == -1 || ans > t)) ans = t;
        t = dijkstra(i, 1);
        if (t != -1 && (ans == -1 || ans > t)) ans = t;
    }
    pt(ans);
    return 0;
}
 

比賽-OBlack學長的訓練賽 (25 Aug, 2018)