Codeforces Round #429(Div 1)

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A. Leha and Function

大家都一眼了结论，只有我个sx真的去证明了…

F(n,k)=1(nk)∑1≤i≤ni(n−ik−1)=−1(nk)∑1≤i≤n((n−i+1)(n−ik−1)−(n+1)(n−ik−1))=−1(nk)⎛⎝∑1≤i≤n(n−i+1)(n−ik−1)−∑1≤i≤n(n+1)(n−ik−1)⎞⎠=−1(nk)⎛⎝k∑1≤i≤n(n−i+1k)−(n+1)∑1≤i≤n(n−ik−1)⎞⎠=−1(nk)⎛⎝k∑i≤n(ik)−(n+1)∑i<n(ik−1)⎞⎠=1(nk)((n+1)(nk)−k(n+1k+1))=1(nk)((n+1)(nk)−k(n+1)k+1(nk))=n+1k+1

猜想：逆序和大于等于乱序和大于等于顺序和。我们只需要证明将B数组排序后，如果存在一对相邻的逆序对，交换后结果不会更劣，再根据冒泡排序算法的正确性即可得证。也即：

n1≤n2,k1≤k2n1+1k1+1+n2+1k2+1≤n2+1k1+1+n1+1k2+1

也就是证明：

n1−n2k1+1≤n1−n2k2+1

也就是证：

1k1+1≥1k2+1

由于k1≤k2，上式显然成立。所以结论就是：逆序和最大。

#include <bits/stdc++.h>using namespace std;const int MAXN = 200005;int A[MAXN], B[MAXN], Ap[MAXN], n;priority_queue<pair<int, int> > pr;priority_queue<pair<int, int>, vector<pair<int,int> >, greater<pair<int, int> > > pr2;int main(){    scanf("%d", &n);    for (int i = 1; i <= n; i++) scanf("%d", &A[i]), pr2.push(make_pair(A[i], i));    for (int i = 1; i <= n; i++) scanf("%d", &B[i]), pr.push(make_pair(B[i], i));    for (int i = 1; i <= n; i++) {        pair<int, int> a = pr2.top(), b = pr.top();        pr.pop(), pr2.pop();        Ap[b.second] = a.first;    }    for (int i = 1; i <= n; i++)        printf("%d ", Ap[i]);    puts("");    return 0;}

B. Leha and another game about graph（补）

考虑增量法构造，每一次选取两个di=1或者一个di=1一个dj=−1，用一条路径连接，路径上的边状态翻转。

正确性在于，如果经过一段已经经过的路径，可以删去这段路径，并将上下两部分视为两个路径，不影响结果。

然后发现任意一个生成森林和整张图是等价的。

然后就可以喜闻乐见的NOIP经典方法树上差分了。由于异或运算的性质，甚至不需要求lca。（当然你也可以选择树剖233）。

#include <bits/stdc++.h>using namespace std;const int MAXN = 300005;int fa[MAXN];inline int findf(int i){ return fa[i]?fa[i]=findf(fa[i]):i; }struct node {    int to, next, id;} edge[MAXN*2];int head[MAXN], top = 0;inline void push(int i, int j, int id){ edge[++top] = (node) {j, head[i], id}, head[i] = top; }int n, m;int d[MAXN];vector<int> vec[MAXN], v[MAXN];int depth[MAXN], pre[MAXN];int tag[MAXN], del[MAXN], cnt = 0;void dfs_calc(int nd, int f){    for (int i = head[nd]; i; i = edge[i].next) {        int to = edge[i].to;        if (to == f) continue;        pre[to] = edge[i].id, dfs_calc(to, nd), tag[nd] ^= tag[to];    }    if (tag[nd] == 1) del[pre[nd]] = 1;}int main(){    scanf("%d%d", &n, &m);    for (int i = 1; i <= n; i++) scanf("%d", &d[i]);    for (int i = 1; i <= m; i++) {        int u, v; scanf("%d%d", &u, &v);        if (findf(u) != findf(v))            fa[findf(u)] = findf(v), push(u, v, i), push(v, u, i);    }    for (int i = 1; i <= n; i++) {        if (d[i] == 1)            vec[findf(i)].push_back(i);        if (d[i] == -1)            v[findf(i)].push_back(i);    }    for (int i = 1; i <= n; i++)        if (fa[i] == 0) {            if ((vec[i].size()&1) && v[i].empty()) { puts("-1"); return 0; }            int k = vec[i].size()/2;            for (int j = 0; j < k; j++) {                int a = vec[i][j], b = vec[i][j+k];                tag[a] ^= 1, tag[b] ^= 1;            }            if (vec[i].size()&1){                int a = vec[i][vec[i].size()-1], b = v[i][0];                tag[a] ^= 1, tag[b] ^= 1;            }            dfs_calc(i, 0);        }    for (int i = 1; i <= n; i++)        if (del[i] == 1)            cnt++;    printf("%d\n", cnt);    for (int i = 1; i <= n; i++)        if (del[i] == 1)            printf("%d ", i);    puts("");    return 0;}

D. Destiny（补）

首先用莫队+set+线段树/树状数组很容易得到O(nn−−√lgn)的算法，但是并不能过…

O(nn−−√lgn)(TLE)

#include <bits/stdc++.h>using namespace std;const int MAXN = 300005, N = 1<<19;int n, q;int a[MAXN];int T[MAXN], ans[MAXN];int bk;struct query {    int x, y, k, id;    friend bool operator < (const query &a, const query &b)    { return a.x/bk == b.x/bk ? a.y < b.y : a.x < b.x; }} qy[MAXN];multiset<int> hp[MAXN];int zkw[N+N+1];inline void modify(int pos, int dt){    if (pos == 0) return;    // cerr << "M" << pos << " " << dt << endl;    pos += N-1;    zkw[pos] = dt;    for (pos >>= 1; pos; pos >>= 1)        zkw[pos] = min(zkw[pos<<1], zkw[pos<<1|1]);}inline int get_ans(int L, int R){    int ans = INT_MAX;    // cerr << L << " " << R << endl;    for (L += N-1, R += N-1; L <= R; L >>= 1, R >>= 1) {        if (L&1) ans = min(ans, zkw[L++]);        if (!(R&1)) ans = min(ans, zkw[R--]);    }    // cerr << "Q" << L << " " << R << " " << ans << endl;    return ans == INT_MAX ? -1 : ans;}inline void pop(int dt){    if (T[dt]) {        hp[T[dt]].erase(dt);        if (!hp[T[dt]].empty()) modify(T[dt], *hp[T[dt]].begin());        else modify(T[dt], INT_MAX);    }    hp[--T[dt]].insert(dt);    modify(T[dt], *hp[T[dt]].begin());}inline void push(int dt){    if (T[dt]) {        hp[T[dt]].erase(dt);        if (!hp[T[dt]].empty()) modify(T[dt], *hp[T[dt]].begin());        else modify(T[dt], INT_MAX);    }    hp[++T[dt]].insert(dt);    modify(T[dt], *hp[T[dt]].begin());}int main(){    scanf("%d%d", &n, &q);    bk = int(sqrt(n)+0.5);    if (!bk) bk = 1;    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);    for (int i = 1; i <= N+N-1; i++) zkw[i] = INT_MAX;    for (int i = 1; i <= q; i++) scanf("%d%d%d", &qy[i].x, &qy[i].y, &qy[i].k), qy[i].id = i;    sort(qy+1, qy+q+1);    int L = 1, R = 0;    for (int i = 1; i <= q; i++) {        while (L < qy[i].x) pop(a[L]), L++;        while (L > qy[i].x) L--, push(a[L]);        while (R < qy[i].y) R++, push(a[R]);        while (R > qy[i].y) pop(a[R]), R--;        int Lb = (qy[i].y-qy[i].x+1)/qy[i].k+1;        ans[qy[i].id] = get_ans(Lb, n);    }    for (int i = 1; i <= q; i++)        printf("%d\n", ans[i]);    return 0;}

考虑分块。设块大小为b，如果序列长度小于b，则直接暴力；序列长度大于b，则如果存在解，这个数字在序列中至少出现b/5次，这样的数不会超过nb/5=5nb个，因此只要预处理出这些数暴力一遍即可。用莫队维护总的复杂度是：

O(qn−−√+q(b+5nb))≥O(qn−−√+2q5n−−√)

等号成立当且仅当b=5nb也就是b=5n−−√。复杂度很不科学但跑得比谁都快…最慢数据只要592ms…

#include <bits/stdc++.h>using namespace std;const int MAXN = 300005;int n, q;int a[MAXN], bk, bk2;struct query {    int x, y, k, id;    friend bool operator < (const query &a, const query &b)    { return a.x/bk2 == b.x/bk2 ? a.y < b.y : a.x < b.x; }} qy[MAXN];int T[MAXN];int g[MAXN], top = 0;int p[MAXN];int main(){    scanf("%d%d", &n, &q);    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), T[a[i]]++;    bk = int(sqrt(5*n)+0.5);    if (bk > n) bk = n;    if (bk < 1) bk = 1;    bk2 = int(sqrt(5*n)+0.5);    if (bk2 > n) bk2 = n;    if (bk2 < 1) bk2 = 1;    for (int i = 1; i <= n; i++)        if (T[i]*5 >= bk)            g[++top] = i;    for (int i = 1; i <= q; i++)        scanf("%d%d%d", &qy[i].x, &qy[i].y, &qy[i].k), qy[i].id = i;    sort(qy+1, qy+q+1);    memset(T, 0, sizeof T);    register int L = 1, R = 0;    for (int i = 1; i <= q; i++) {        while (L < qy[i].x) T[a[L++]]--;        while (L > qy[i].x) T[a[--L]]++;        while (R > qy[i].y) T[a[R--]]--;        while (R < qy[i].y) T[a[++R]]++;        int ans = INT_MAX, cnt = (qy[i].y-qy[i].x+1)/qy[i].k+1;        if (qy[i].y-qy[i].x+1 <= bk) {            for (register int j = qy[i].x; j <= qy[i].y; j++)                if (T[a[j]] >= cnt)                    ans = min(ans, a[j]);        } else {            for (register int j = 1; j <= top; j++)                if (T[g[j]] >= cnt)                    ans = min(ans, g[j]);        }        p[qy[i].id] = ans==INT_MAX?-1:ans;    }    for (int i = 1; i <= q; i++)        printf("%d\n", p[i]);    return 0;}