[APIO2015]巴邻旁之桥

Bzoj权限题

luogu题面

先去掉同边的

首先k==1，即求一个点j

使\(\sum_{i\in A} |D_i - D_j| + \sum_{i\in B} |D_i - D_j|\)最小

因为两边j是一样的，直接合在一起就好

所以就是\(\sum |D_i - D_j|\)最小

那么j就是的中位数，合在一起排序就好了

然后k==2，

设每个人的两个位置为A，B那么他肯定去离\(\frac{A+B}{2}\)最近的桥

所以可以先把这些人按\(\frac{A+B}{2}\)排序，再枚举这些人从哪里分开，左右单独考虑取中位数就好，最后取min

也就是说现在要动态维护区间中位数，可以用线段树或者平衡树维护size，排序后插入，树上二分找中位数，比较左右size即可

# include <bits/stdc++.h>

# define IL inline

# define RG register

# define Fill(a, b) memset(a, b, sizeof(a))

# define Copy(a, b) memcpy(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(2e5 + 10), __(1e6 + 10);

IL ll Read(){

    RG char c = getchar(); RG ll x = 0, z = 1;

    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;

    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);

    return x * z;

}

int k, n, cnt, o[_], len, sz[2][__];

struct Zsy{

	int a, b;

	IL bool operator <(RG Zsy Orz) const{  return a + b < Orz.a + Orz.b;  }

} p[_];

ll ans, sum[2][__];

IL void Modify(RG int c, RG int x, RG int l, RG int r, RG int v, RG int op){

	sz[c][x] += op; sum[c][x] += op * o[v];

	if(l == r) return;

	RG int mid = (l + r) >> 1;

	if(v <= mid) Modify(c, x << 1, l, mid, v, op);

	else Modify(c, x << 1 | 1, mid + 1, r, v, op);

}

IL int Find(RG int c, RG int x, RG int l, RG int r, RG int k){

	if(l == r) return l;

	RG int mid = (l + r) >> 1;

	if(k <= sz[c][x << 1]) return Find(c, x << 1, l, mid, k);

	return Find(c, x << 1 | 1, mid + 1, r, k - sz[c][x << 1]);

}

IL ll Query(RG int c, RG int x, RG int l, RG int r, RG int L, RG int R, RG int op){

	if(L <= l && R >= r) return op ? sum[c][x] : sz[c][x];

	RG int mid = (l + r) >> 1; RG ll yyb = 0;

	if(L <= mid) yyb = Query(c, x << 1, l, mid, L, R, op);

	if(R > mid) yyb += Query(c, x << 1 | 1, mid + 1, r, L, R, op);

	return yyb;

}

int main(RG int argc, RG char* argv[]){

	k = Read(); n = Read();

	if(k == 1){

		for(RG int i = 1; i <= n; ++i){

			RG char c1, c2; RG int ss, tt;

			scanf(" %c", &c1); ss = Read();

			scanf(" %c", &c2); tt = Read();

			if(c1 == c2) ans += abs(ss - tt);

			else o[++cnt] = ss, o[++cnt] = tt;

		}

		RG int mid = cnt >> 1; ans += mid;

		sort(o + 1, o + cnt + 1);

		for(RG int i = 1; i <= cnt; ++i) ans += abs(o[i] - o[mid]);

		printf("%lld\n", ans);

	}

	else{

		RG ll ret = 0; ans = 1e18;

		for(RG int i = 1; i <= n; ++i){

			RG char c1, c2; RG int ss, tt;

			scanf(" %c", &c1); ss = Read();

			scanf(" %c", &c2); tt = Read();

			if(c1 == c2) ret += abs(ss - tt);

			else p[++cnt].a = ss, p[cnt].b = tt, o[++len] = ss, o[++len] = tt;

		}

		ret += cnt;	sort(p + 1, p + cnt + 1);

		sort(o + 1, o + len + 1); len = unique(o + 1, o + len + 1) - o - 1;

		RG ll ss1 = 0, ss2 = 0;

		for(RG int i = 1; i <= cnt; ++i){

			ss2 += p[i].a + p[i].b;

			p[i].a = lower_bound(o + 1, o + len + 1, p[i].a) - o;

			p[i].b = lower_bound(o + 1, o + len + 1, p[i].b) - o;

			Modify(1, 1, 1, len, p[i].a, 1); Modify(1, 1, 1, len, p[i].b, 1);

		}

		for(RG int i = 1; i <= cnt; ++i){

			ss1 += o[p[i].a] + o[p[i].b]; ss2 -= o[p[i].a] + o[p[i].b];

			Modify(0, 1, 1, len, p[i].a, 1); Modify(0, 1, 1, len, p[i].b, 1);

			Modify(1, 1, 1, len, p[i].a, -1); Modify(1, 1, 1, len, p[i].b, -1);

			RG int mid1 = Find(0, 1, 1, len, i), mid2 = Find(1, 1, 1, len, cnt - i);

			RG ll s1 = Query(0, 1, 1, len, 1, mid1, 1), s2 = Query(1, 1, 1, len, 1, mid2, 1);

			RG ll z1 = Query(0, 1, 1, len, 1, mid1, 0), z2 = Query(1, 1, 1, len, 1, mid2, 0);

			RG ll tot1 = z1 * o[mid1] - s1 + (ss1 - s1) - (2 * i - z1) * o[mid1];

			RG ll tot2 = z2 * o[mid2] - s2 + (ss2 - s2) - (2 * (cnt - i) - z2) * o[mid2];

			ans = min(ans, tot1 + tot2 + ret);

		}

		printf("%lld\n", cnt ? ans : ret);

	}

    return 0;

}

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[APIO2015]巴邻旁之桥

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