P2824 HEOI2016 排序

P2824

很妙的一个题

考虑简化版：将一个01序列排序？(logn)复杂度

线段树维护，查询区间内\(1\)的个数记为\(cnt1\)，升序，将\([r-cnt1+1,r]\)改为\(1\)，将\([l,r-cnt1]\)改为\(0\)，降序将\([l,l+cnt1-1]\)改为\(1\),将\([l+cnt,r]\)改为\(0\)，成功将排序转化为区间查询和区间修改

离线

此题二分答案\(mid\)，把原序列中大等于\(mid\)标记为\(1\)，小于\(mid\)标记为\(0\)，对于每个操作将\(01\)序列排序，如果第\(p\)个位子仍是\(1\)的话就是可行的

如果二分一个值成立当且仅当这个位子的值大于等于\(mid\)，故如果\(check\)返回\(true\)，则\(l = mid+1\)，否则\(r = mid-1\)，

#include<cstdio>
#include<cstring>
#include<cctype>
#define lc o<<1
#define rc o<<1|1
#define mid (l+r)/2
using namespace std;

const int N = 100010;
int n, m, p;
int T[4*N], lazy[4*N];//segment tree
int a[N], ch[N], L[N], R[N];//the information by reading

inline int read()
{
    char ch = getchar(); int x = 0;
    while (!isdigit(ch)) ch = getchar();
    while (isdigit(ch)){ x = x*10+ch-'0'; ch = getchar(); }
    return x;
}

inline void build(int o, int l, int r, int x)
{
    if (l == r){
        T[o] = a[l] >= x;
        lazy[o] = 0;
        return;
    }
    build(lc, l, mid, x); build(rc, mid+1, r, x);
    T[o] = T[lc]+T[rc]; lazy[o] = 0;
}

inline void pushdown(int o, int l, int r)
{
    if (!lazy[o]) return;
    lazy[lc] = lazy[rc] = lazy[o];
    if (lazy[o] == 1){
        T[lc] = mid-l+1; T[rc] = r-mid;
    } else T[lc] = T[rc] = 0;
    lazy[o] = 0;
}

inline int query(int o, int l, int r, int x, int y)
{
    if (x <= l && y >= r) return T[o];
    if (x > r || y < l) return 0;
    pushdown(o, l, r);
    return query(lc, l, mid, x, y) + query(rc, mid+1, r, x, y);
}

inline int queryPoint(int o, int l, int r, int x)
{
    if (l == x && r == x) return T[o];
    pushdown(o, l, r);
    if (x <= mid) return queryPoint(lc, l, mid, x);
    else return queryPoint(rc, mid+1, r, x);
}

inline void update(int o, int l, int r, int x, int y, int val)
{
    if (x <= l && y >= r){
        T[o] = val*(r-l+1); lazy[o] = val ? 1 : -1;
        return;
    }
    if (x > r || y < l) return;
    pushdown(o, l, r);
    update(lc, l, mid, x, y, val);
    update(rc, mid+1, r, x, y, val);
    T[o] = T[lc]+T[rc];
}

inline bool check(int x)
{
    build(1, 1, n, x);
    for (int i = 1; i <= m; i ++){
        int cnt1 = query(1, 1, n, L[i], R[i]);
        if (ch[i] == 0){
            update(1, 1, n, R[i]-cnt1+1, R[i], 1);
            update(1, 1, n, L[i], R[i]-cnt1, 0);
        }
        else{
            update(1, 1, n, L[i], L[i]+cnt1-1, 1);
            update(1, 1, n, L[i]+cnt1, R[i], 0);
        }
    }
    return queryPoint(1, 1, n, p);
}

int main()
{
    n = read(); m = read();
    for (int i = 1; i <= n; i ++) a[i] = read();
    for (int i = 1; i <= m; i ++){
        ch[i] = read(); L[i] = read(); R[i] = read();
    }
    p = read();
    int ll = 1, rr = n, midd, ans;
    while (ll <= rr){
        midd = (ll+rr) >> 1;
        if (check(midd)) ans = midd, ll = midd+1; else rr = midd-1;
    }
    printf("%d\n", ans);
    return 0;
}