luoguP3415 祭坛
Posted ljc00118
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https://www.luogu.org/problemnew/show/P3415
考虑二分结界层数,将 n 个点按 x 大小依次加入答案,一行一行的做,用树状数组维护当前这一行中[0, x - 1] 包含祭坛大于 mid 的且 [x + 1, n] 中包含的祭坛也大于 mid 的坐标,再计算出这一行有几个地方可以作为中心,简单容斥(可能还不算容斥?)一下就可以了
有一个坑就是当节点层数是 0 的时候输出两行 0
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
struct ele {
int x, y;
bool operator < (const ele A) const {return x < A.x || (x == A.x && y < A.y);}
}d[N];
int sum[N], f[N], y[N], allsum[N];
int n, len;
inline int lowbit(int x) {return x & -x;}
inline void add(int x, int y) {for(int i = x; i <= n; i += lowbit(i)) f[i] += y;}
inline int query(int x) {int ans = 0; for(int i = x; i; i -= lowbit(i)) ans += f[i]; return ans;}
int solve(int mid) {
memset(sum, 0, sizeof(sum));
memset(f, 0, sizeof(f));
int now = 1, ans = 0;
for(int i = 0; i <= n; i++) {
len = 0;
while(now <= n && d[now].x == i) {
int t = d[now++].y;
y[++len] = t;
}
if(len >= (mid << 1)) {
int l = y[mid] + 1, r = y[len - mid + 1] - 1;
for(int j = mid + 1; j <= len - mid; j++) ans -= (sum[y[j]] >= mid && (allsum[y[j]] - sum[y[j]] >= mid));
ans += (query(r) - query(l - 1));
}
for(int j = 1; j <= len; j++) {
int t = y[j];
if(sum[t] >= mid && (allsum[t] - sum[t] == mid)) add(t, -1);
sum[t]++;
if(sum[t] == mid && (allsum[t] - sum[t] >= mid)) add(t, 1);
}
}
return ans;
}
int main() {
cin >> n;
for(int i = 1; i <= n; i++) scanf("%d %d", &d[i].x, &d[i].y);
sort(d + 1, d + n + 1);
for(int i = 1; i <= n; i++) allsum[d[i].y]++;
int l = 1, r = n / 4;
while(l < r) {
int mid = (l + r + 1) >> 1;
if(solve(mid)) l = mid;
else r = mid - 1;
}
if(solve(l) == 0) printf("0
0
");
else printf("%d
%d
", l, solve(l));
return 0;
}
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