NOIP模拟 Work - 二分 + 树状数组 / ???

题目分析

如果没有最后的注意事项，此题就是二分裸题。有了注意事项，会有两种思路:

• 在线：二分天数t，并在主席树上求1~t天中大于d(浪费的时间)的时间之和以及数量，答案即为：sum - d * cnt 无奈写的丑，卡卡只能过6、7个点。
• 离线：简单考虑，既然要求大于等于d的和以及数量，不妨按照d来排序，再把t也排序。每次将大于等于d的t加入树状数组（记录和以及数量），这样就能直接查出和和数量。 AC。

code

树状数组 AC

#include<bits/stdc++.h>
using namespace std;

const int N = 200005, M = 1000005;
typedef long long ll;
int n, m, anss[N];
int pointer;
typedef pair<int, int> P;
typedef pair<ll, int> ansP;
P t[N];

struct BIT{
	ll sum[N], cnt[N];
	inline void add(int x, ll v, int c){
		for(int i = x; i <= m; i += (i&-i))
			sum[i] += v, cnt[i] += c;
	}
	inline ansP query(int x){
		ansP ret = P(0, 0);
		for(int i = x; i; i -= (i&-i))
			ret.first += sum[i], ret.second += cnt[i];
		return ret;
	}
}bit;

struct node{
	int d, r, id;
	inline bool operator < (const node &b) const{
		return d < b.d;
	}
}s[N];

inline int read(){
    int i = 0, f = 1; char ch = getchar();
    for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
    if(ch == '-') f = -1, ch = getchar();
    for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch -'0');
    return i * f;
}

inline void wr(int x){
	if(x < 0) putchar('-'), x = -x;
	if(x > 9) wr(x / 10);
	putchar(x % 10 + '0');
}

inline bool calc(int mid, int d, int r){
    ansP tmp = bit.query(mid);
    ll ret = tmp.first - 1LL * d * tmp.second;
    return ret >= r;
}

int main(){
	n = read(), m = read();
	for(int i = 1; i <= m; i++) t[i].first = read(), t[i].second = i;
	for(int i = 1; i <= n; i++) s[i].d = read(), s[i].r = read(), s[i].id = i;
	sort(t + 1, t + m + 1), sort(s + 1, s + n + 1);
	pointer = m;
	for(int i = n; i >= 1; i--){
		while(t[pointer].first >= s[i].d && pointer >= 1) bit.add(t[pointer].second, t[pointer].first, 1), pointer--;
		int l = 1, r = m, ans = 0;
		while(l <= r){
			int mid = l + r >> 1;
			if(calc(mid, s[i].d, s[i].r)) ans = mid, r = mid - 1;
			else l = mid + 1;
		}
		anss[s[i].id] = ans;
	}
	for(int i = 1; i <= n; i++) wr(anss[i]), putchar(' ');
	return 0;
}

主席树 50 ~ 60

#include<bits/stdc++.h>
using namespace std;

const int N = 200005, M = 1000005;
typedef long long ll;

struct node{
    node *lc, *rc;
    ll sum, cnt;
    inline void upt(){
    	sum = lc->sum + rc->sum;
    	cnt = lc->cnt + rc->cnt;
	}
}pool[N * 20], *tail = pool, *null = pool, *rt[N];
int maxx = -1;
typedef pair<ll, ll> ansP;

int n, m, d[N], rr[N];
int t[N], b[N * 2], len;
ll tsum[N];

inline int read(){
    int i = 0, f = 1; char ch = getchar();
    for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
    if(ch == '-') f = -1, ch = getchar();
    for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch -'0');
    return i * f;
}

inline void wr(int x){
	if(x < 0) putchar('-'), x = -x;
	if(x > 9) wr(x / 10);
	putchar(x % 10 + '0');
}

inline void insert(node *x, node *&y, int l, int r, int v){
    y = ++tail;
    y->lc = x->lc, y->rc = x->rc;
    y->sum = x->sum, y->cnt = x->cnt;
    y->sum += b[v];
    y->cnt++;
    if(l == r) return;
    int mid = l + r >> 1;
    if(v <= mid) insert(x->lc, y->lc, l, mid, v);
    else insert(x->rc, y->rc, mid + 1, r, v);
    y->upt();
}

inline ansP query(node *nl, node *nr, int l, int r, int x, int y){
    if(x <= l && r <= y) return ansP(nr->sum - nl->sum, nr->cnt - nl->cnt);
    int mid = l + r >> 1;
    ansP ret = ansP(0, 0);
    if(x <= mid){
        ansP tmp = query(nl->lc, nr->lc, l, mid, x, y);
        ret.first += tmp.first;
        ret.second += tmp.second;
    }
    if(y > mid){
        ansP tmp = query(nl->rc, nr->rc, mid + 1, r, x, y);
        ret.first += tmp.first;
        ret.second += tmp.second;
    }
    return ret;
}

bool flag;
inline bool calc(int mid, int d, int r){
    ansP tmp = query(rt[0], rt[mid], 1, maxx, d, maxx);
    ll ret = tmp.first - 1LL * b[d] * tmp.second;
    return ret >= r;
}

inline void disc_init(){
    sort(b + 1, b + len + 1);
    len = unique(b + 1, b + len + 1) - (b + 1);
    for(int i = 1; i <= m; i++) t[i] = lower_bound(b + 1, b + len + 1, t[i]) - b;
    for(int i = 1; i <= n; i++) d[i] = lower_bound(b + 1, b + len + 1, d[i]) - b;
}

int main(){
	freopen("h.in", "r", stdin);
	ios::sync_with_stdio(false);
	cin.tie(NULL), cout.tie(NULL);
    n = read(), m = read();
    null->lc = null->rc = null, null->sum = null->cnt = 0;
    for(int i = 1; i <= m; i++) t[i] = b[++len] = read(), maxx = max(maxx, t[i]);
    for(int i = 1; i <= n; i++) d[i] = b[++len] = read(), rr[i] = read();
    disc_init();
    rt[0] = null;
    for(int i = 1; i <= m; i++)
        insert(rt[i - 1], rt[i], 1, maxx, t[i]);
    for(int i = 1; i <= n; i++){
        int l = 1, r = m, ans = 0;
        if(d[i] > maxx){
			putchar('0'),putchar(' ');
			continue;
		}
        while(l <= r){
            int mid = l + r >> 1;
            if(calc(mid, d[i], rr[i])) ans = mid, r = mid - 1;
            else l = mid + 1;
        }
        cout << ans << " ";
    }
}