省队集训 Day5 选举

【题目大意】

小奇和魔法猪要竞选膜钟国的总统。

有 $n$ $n$

现在你需要把选民分成若干个区间，每个区间的长度在$[l,r]$ $[l, r]$

小奇想要赢得选举，于是它请你合理地分区间，使得它获得的票数减去魔法猪获得的票数最大。如果你帮它解决了这个问题，它会送你这道题的题解。

$n leq 10^6$

【题解】

考虑暴力dp，复杂度为$O(n^2)$。

然后考虑优化暴力，每次相当于区间往右移动一格，相当于加入1个数，减去1个数，这个可以用线段树来维护

如果普通写法需要开单调队列，我们可以把下标映射到位置，这样就能执行减去1个数的操作了。

复杂度$O(nlogn)$

# include <vector>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

const int M = 5e5 + 10, N = 1e6 + 10, inf = 1e9, SN = 1024 * 1024 + 5;

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

int n, L, R, a[N], s[N], f[N], pos[N], idx;
vector<int> v[N + N]; int l[N], r[N];

struct SMT {
    int w[N << 2];
    # define ls (x<<1)
    # define rs (x<<1|1)
    inline void up(int x) {
        w[x] = max(w[ls], w[rs]);
    }
    inline void build(int x, int l, int r) {
        w[x] = -inf;
        if(l == r) return ;
        int mid = l+r>>1;
        build(ls, l, mid); build(rs, mid+1, r);
    }
    inline void edt(int x, int l, int r, int pos, int del) {
        if(l == r) {
            w[x] = del;
            return ;
        }
        int mid = l+r>>1;
        if(pos <= mid) edt(ls, l, mid, pos, del);
        else edt(rs, mid+1, r, pos, del);
        up(x);
    }
    inline int query(int x, int l, int r, int L, int R) {
        if(L > R) return -inf;
        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 max(query(ls, l, mid, L, mid), query(rs, mid+1, r, mid+1, R));
    }
    # undef ls
    # undef rs
}T;

inline void ins(int i) {
//    cout << "add " << pos[i] << endl;
    T.edt(1, 0, n, pos[i], f[i]);
}

inline void del(int i) {
    T.edt(1, 0, n, pos[i], -inf);
}

int main() {
    freopen("election.in", "r", stdin);
    freopen("election.out", "w", stdout);
    n = getint(), L = getint(), R = getint();
    for (int i=1; i<=n; ++i) a[i] = getint();
    for (int i=1; i<=n; ++i) s[i] = s[i-1] + a[i];
//    for (int i=0; i<=n; ++i) cout << s[i] << ' '; cout << endl;
    
    for (int i=0; i<=n; ++i) v[s[i] + n].push_back(i);
    for (int i=0; i<=n+n; ++i) {
        int lst = idx;
        for (int j=0; j<v[i].size(); ++j) 
            pos[v[i][j]] = idx++;
        for (int j=0; j<v[i].size(); ++j)
            l[v[i][j]] = lst, r[v[i][j]] = idx - 1;
    }
//    for (int i=0; i<=n; ++i) cout << pos[i] << ' ' << l[i] << ' ' << r[i] << endl; cout << endl;
    
    f[0] = 0;
    T.build(1, 0, n);
    for (int i=1; i<=n; ++i) {
        if(i - R - 1 >= 0) del(i - R - 1);
        if(i - L >= 0) ins(i - L);
        int ta = T.query(1, 0, n, 0, l[i]-1), tb = T.query(1, 0, n, r[i]+1, n), tc = T.query(1, 0, n, l[i], r[i]);
        if(ta != -inf) ta ++;
        if(tb != -inf) tb --;
//        cout << i << "???" << l[i]-1 <<endl;
//        cout << i << ' ' << ta << ' ' << tb << ' ' << tc << endl;
        f[i] = max(ta, max(tb, tc));
    }
        
//    }
    if(f[n] == -inf) cout << "Impossible";
    else cout << f[n];
    return 0;
}

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