[BJOI2017]开车

P3992

题目大意

给出 \(n\) 个车的位置 \(a_i\) 和 \(n\) 个加油站的位置 \(b_i\) ，一辆车从 \(x\) 位置到 \(y\) 位置的代价为 \(|x - y|\) ，问如何两两配对使答案最小，\(q\) 次询问每次修改 \(a_x\) 为 \(y\) ，每次修改后给出当前答案。

数据范围

\(n \le 50000, q \le 50000\)

时空限制

3000ms, 256MB, O2

反思

似乎根本没有任何思考就点开了题解

分析

首先发现答案实际上就是分别排序后对应配对，考虑维护数轴上每条边的贡献，发现我们记一个 \(a_i\) 为 \(1\) ，$b_i $ 为 \(-1\) 的前缀和，那么它的贡献就是 \(|sum|\)

那么我们修改就相当于区间 \(\pm 1\) ，但是我们无法简单地维护一个绝对值，所以用分块的方法

将块内的前缀和排序，二分 \(0\) 点，带上标记即可修改

Code

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
inline char nc() {static char buf[100000], *l = buf, *r = buf;return l==r&&(r=(l=buf)+fread(buf,1,100000,stdin),l==r)?EOF:*l++;
}
template<class T> void read(T &x) {x = 0; int f = 1, ch = nc();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=nc();}while(ch>='0'&&ch<='9'){x=x*10-'0'+ch;ch=nc();}x *= f;
}
#define myabs(a) ((a) < 0 ? - (a) : (a))
typedef long long ll;
const int maxn = 50000 + 5;
const int maxq = 50000 + 5;
const int maxp = maxn * 2 + maxq;
const int maxb = 400;
int n, q, a[maxn], b[maxn]; ll an;
int H[maxp], p;
struct data {int x, y;
} ask[maxq];
void Hash() {sort(H + 1, H + p + 1);p = unique(H + 1, H + p + 1) - H - 1;for(int i = 1; i <= n; ++i) {a[i] = lower_bound(H + 1, H + p + 1, a[i]) - H;}for(int i = 1; i <= n; ++i) {b[i] = lower_bound(H + 1, H + p + 1, b[i]) - H;}for(int i = 1; i <= q; ++i) {ask[i].y = lower_bound(H + 1, H + p + 1, ask[i].y) - H;}
}
namespace blocks {int len;int L[maxb], R[maxb], bel[maxp];int sum[maxp], val[maxp];int sval[maxp], ord[maxp], tag[maxb];inline int cmp(const int &a, const int &b) {return sum[a] < sum[b];} void rebuild(int x) {sort(ord + L[x], ord + R[x] + 1, cmp);sval[L[x]] = val[ord[L[x]]];for(int i = L[x] + 1; i <= R[x]; ++i) {sval[i] = sval[i - 1] + val[ord[i]];}}void init() {for(int i = 1; i <= n; ++i) {sum[a[i]]++, sum[b[i]]--;}for(int i = 1; i <= p; ++i) {if(i < p) val[i] = H[i + 1] - H[i];sum[i] += sum[i - 1];an += (ll)myabs(sum[i]) * val[i];}len = ceil(sqrt(p));for(int i = 1; i <= p; ++i) {int x = bel[i] = (i - 1) / len + 1;if(!L[x]) L[x] = i;R[x] = i;}for(int i = 1; i <= p; ++i) ord[i] = i;for(int i = 1; i <= bel[p]; ++i) {rebuild(i);}}void add(int x) {for(int i = x; i <= R[bel[x]]; ++i) {an += val[i] * (sum[i] + tag[bel[x]] >= 0 ? 1 : -1);sum[i]++;}rebuild(bel[x]);for(int i = bel[x] + 1; i <= bel[p]; ++i) {int l = L[i], r = R[i], re = -1;while(l <= r) {int mid = (l + r) >> 1;if(sum[ord[mid]] + tag[i] >= 0) r = mid - 1, re = mid;else l = mid + 1;}if(re == -1) {an -= sval[R[i]];}else if(re == L[i]) {an += sval[R[i]];}else {an -= sval[re - 1];an += sval[R[i]] - sval[re - 1];}tag[i]++;}}void sub(int x) {for(int i = x; i <= R[bel[x]]; ++i) {an += val[i] * (sum[i] + tag[bel[x]] <= 0 ? 1 : -1);sum[i]--;}rebuild(bel[x]);for(int i = bel[x] + 1; i <= bel[p]; ++i) {int l = L[i], r = R[i], re = -1;while(l <= r) {int mid = (l + r) >> 1;if(sum[ord[mid]] + tag[i] <= 0) l = mid + 1, re = mid;else r = mid - 1;}if(re == -1) {an -= sval[R[i]];}else if(re == R[i]) {an += sval[R[i]];}else {an += sval[re];an -= sval[R[i]] - sval[re];}tag[i]--;}}
}
int main() {
//  freopen("testdata.in", "r", stdin);
//  freopen("testdata.out", "w", stdout); read(n);for(int i = 1; i <= n; ++i) {read(a[i]), H[++p] = a[i];}for(int i = 1; i <= n; ++i) {read(b[i]), H[++p] = b[i];}read(q); for(int i = 1; i <= q; ++i) {read(ask[i].x), read(ask[i].y);H[++p] = ask[i].y;}Hash();blocks :: init();printf("%lld\n", an);for(int i = 1; i <= q; ++i) {int x = ask[i].x, y = ask[i].y;blocks :: sub(a[x]);blocks :: add(a[x] = y);printf("%lld\n", an);}return 0;
}

总结

利用分块维护绝对值