POJ

洛谷

分析

離散化+前綴和+二分

這題和激光炸彈很像，但由于坐標(biāo)范圍較大，需要用到二分。

代碼

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define il inline
#define maxn 10007
#define re register
#define tie0 cin.tie(0),cout.tie(0)
#define fastio ios::sync_with_stdio(false)
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
typedef long long ll;template <typename T> inline void read(T &x) {T f = 1; x = 0; char c;for (c = getchar(); !isdigit(c); c = getchar()) if (c == '-') f = -1;for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);x *= f;
}struct pos {int x, y;
}p[maxn>>1];int n, c, cx, cy;
int x[maxn>>1], y[maxn>>1], hx[maxn], hy[maxn], sum[maxn>>1][maxn>>1];bool check(int lim) {for (int i = hx[lim]; i <= cx; ++i)for (int j = hy[lim]; j <= cy; ++j) {int x0 = 0, y0  = 0;if (x[i] - lim >= 0) x0 = hx[x[i]-lim];if (y[j] - lim >= 0) y0 = hy[y[j]-lim];if (sum[i][j] + sum[x0][y0] - sum[i][y0] - sum[x0][j] >=c ) return 1;}return 0;
}int main() {read(c), read(n);for (int i = 1; i <= n; ++i) {read(p[i].x), read(p[i].y);hx[p[i].x]++, hy[p[i].y]++;}for (int i = 1; i <= 10000; ++i) {if (hx[i]) x[++cx] = i; hx[i] = cx;if (hy[i]) y[++cy] = i; hy[i] = cy;}for (int i = 1; i <= n; ++i) sum[hx[p[i].x]][hy[p[i].y]]++;for (int i = 1; i <= cx; ++i)for (int j = 1; j <= cy; ++j)sum[i][j] += sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1];int l = 1, r = 10000;while (l < r) {int mid = l + r >> 1;if (check(mid)) r = mid;else l = mid + 1;}printf("%d", l);return 0;
}

轉(zhuǎn)載于:https://www.cnblogs.com/hlw1/p/11409427.html

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