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 << " ";
}
}
