省队集训Day1 总统选举

【题目大意】

一个$n$个数的序列，$m$次操作，每次选择一段区间$[l, r]$，求出$[l, r]$中出现超过一半的数。

如果没有超过一半的数，那么就把答案钦定为$s$，每次会有$k$个数进行改变，给出下标，改变成当前的答案$s$。

$n, m leq 5*10^5, sum kleq 10^6$

By FJSDFZ ditoly

【题解】

用这题的方法进行线段树操作即可：http://www.cnblogs.com/galaxies/p/20170602_c.html

但是这样需要验证是否可行，那么问题变成待修改的询问$[l, r]$区间内$x$的出现次数。

如果没有修改，我会分块！

修改就写个平衡树（开始想STL发现set不支持iterator减法，就gg了）

然后慢的要死。（论不会treap的危害）

后来一气之下找了pb_ds::tree来用。

# include <ext/pb_ds/assoc_container.hpp>
# include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
tree <int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> S[M];

每次只要调用order_of_key(x)即可，注意这个操作是右开区间，所以本题中可能需要些许变换。

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10;
const int mod = 1e9+7;

inline int getint() {
    int x = 0; char ch = getchar();
    while(!isdigit(ch)) ch = getchar();
    while(isdigit(ch)) x = (x<<3) + (x<<1) + ch - '0', ch = getchar();
    return x;
}

# define gi getint()

int n, a[M];

struct node {
    int w, t;
    node() {}
    node(int w, int t) : w(w), t(t) {}
    friend node operator + (node a, node b) {
        if(a.w == b.w) return node(a.w, a.t + b.t);
        if(a.t >= b.t) return node(a.w, a.t - b.t);
        else return node(b.w, b.t - a.t);
    }
};

/*
int rt[M];
struct Splay {
    int ch[M][2], fa[M], val[M], s[M];
    inline void up(int x) {
        if(!x) return ;
        s[x] = 1 + s[ch[x][0]] + s[ch[x][1]];
    }
    inline void rotate(int x, int &rt) {
        int y = fa[x], z = fa[y], ls = ch[y][1] == x, rs = ls^1;
        if(rt == y) rt = x;
        else ch[z][ch[z][1] == y] = x;
        fa[ch[x][rs]] = y, fa[y] = x, fa[x] = z;
        ch[y][ls] = ch[x][rs]; ch[x][rs] = y;
        up(y), up(x);
    }
    inline void splay(int x, int &rt) {
        while(x != rt) {
            int y = fa[x], z = fa[y];
            if(y != rt) {
                if((ch[z][0] == y) ^ (ch[y][0] == x)) rotate(x, rt);
                else rotate(y, rt);
            }
            rotate(x, rt);
        }
    }
    
    inline int find(int x, int pos) {
        if(pos == val[x]) return x;
        if(pos < val[x]) return find(ch[x][0], pos);
        else return find(ch[x][1], pos);    
    }
    
    inline int gmax(int x) {
        while(ch[x][1]) x = ch[x][1];
        return x;
    }
    
    inline int del(int &rt, int pos) {
        int x = find(rt, pos);
        splay(x, rt);
        if(!ch[x][0]) {
            rt = ch[x][1]; fa[rt] = 0; ch[x][1] = 0;
            return x;
        }
        if(!ch[x][1]) {
            rt = ch[x][0]; fa[rt] = 0; ch[x][0] = 0;
            return x;
        }
        int y = gmax(ch[x][0]); splay(y, ch[x][0]);
        rt = y; ch[y][1] = ch[x][1]; fa[ch[x][1]] = y;
        ch[x][0] = ch[x][1] = 0; fa[y] = 0; 
        up(y);
        return x;
    }
        
    inline void ins(int &x, int y, int pos, int id) {
        if(!x) {
            x = id; fa[x] = y; val[x] = pos; s[x] = 1; ch[x][0] = ch[x][1] = 0;
            return ;
        }
        if(pos < val[x]) ins(ch[x][0], x, pos, id);
        else ins(ch[x][1], x, pos, id);
        up(x);
    }
    
    inline void Ins(int &rt, int pos, int id) {
        ins(rt, 0, pos, id);
        splay(id, rt);
    }
    
    inline int rank(int x, int d) {
        if(!x) return 0;
        if(d < val[x]) return rank(ch[x][0], d);
        else return s[ch[x][0]] + 1 + rank(ch[x][1], d);
    }
    
    inline void DEBUG(int x) {
        if(!x) return ;
        DEBUG(ch[x][0]);
        printf("x = %d, ls = %d, rs = %d, pos = %d, sz = %d
", x, ch[x][0], ch[x][1], val[x], s[x]);
        DEBUG(ch[x][1]);
    }
    
    inline void debug() {
        for (int i=1; i<=5; ++i) {
            printf("rt %d ======================
", i);
            DEBUG(rt[i]);
        }
        puts("
");
    }
}S;
*/


# include <ext/pb_ds/assoc_container.hpp>
# include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
tree <int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> S[M];

struct SMT {
    node w[M << 2];
    # define ls (x<<1)
    # define rs (x<<1|1)
    inline void up(int x) {
        w[x] = w[ls] + w[rs];
    }
    inline void build(int x, int l, int r) {
        if(l == r) {
            w[x] = node(a[l], 1);
            return ;
        }
        int mid = l+r>>1;
        build(ls, l, mid), build(rs, mid+1, r);
        up(x);
    }
    inline void edt(int x, int l, int r, int ps, int d) {
        if(l == r) {
            S[w[x].w].erase(l);
            w[x] = node(d, 1); 
            S[d].insert(l);
            return ;
        }
        int mid = l+r>>1;
        if(ps <= mid) edt(ls, l, mid, ps, d);
        else edt(rs, mid+1, r, ps, d);
        up(x);
    }
    inline node query(int x, int l, int r, int L, int R) {
        if(L <= l && r <= R) return w[x];
        int mid = l+r>>1;
        if(R <= mid) return query(ls, l, mid, L, R);
        else if(L > mid) return query(rs, mid+1, r, L, R);
        else return query(ls, l, mid, L, mid) + query(rs, mid+1, r, mid+1, R);
    }
    inline void debug(int x, int l, int r) {
        printf("x = %d, [%d, %d],    w[x] = {%d, %d}
", x, l, r, w[x].w, w[x].t);
        if(l == r) return ;
        int mid = l+r>>1;
        debug(ls, l, mid);
        debug(rs, mid+1, r);
    }
    # undef ls
    # undef rs
}T;
    
inline bool check(int l, int r, int d) {
    int times = S[d].order_of_key(r+1) - S[d].order_of_key(l);
//    cout << "times = " << times << endl;
    if(times <= (r-l+1)/2) return false;
    return true;
}

int main() {
    freopen("president.in", "r", stdin);
    freopen("president.out", "w", stdout);
    n = gi; int Q = gi; 
    for (int i=1; i<=n; ++i) {
        a[i] = gi;
        S[a[i]].insert(i);
    }
    T.build(1, 1, n);
    int l, r, s, k, leader; node ans;
    while(Q--) {
        if(Q % 10000 == 0) cerr << Q << endl;
//        S.debug();
        l = gi, r = gi, s = gi, k = gi;
        ans = T.query(1, 1, n, l, r);
        if(check(l, r, ans.w)) leader = ans.w;
        else leader = s;
        for (int i=1, t; i<=k; ++i) {
            t = gi;
            T.edt(1, 1, n, t, leader);
        }
        printf("%d
", leader);
//        T.debug(1, 1, n);
    }
    ans = T.query(1, 1, n, 1, n);
    if(check(1, n, ans.w)) printf("%d
", ans.w);
    else puts("-1");
    return 0;
}

注释里面是平衡树。。T得不要不要的。。

复杂度都是$O((n + m + sum K) log n)$