「笔记」二分与三分

最近老师让我们重新开始学一本通提高篇，虽然这本书比较烂，但是由于我很菜，什么都不会，所以必须（\(lhy\)大佬喜欢说必须）好好学！还有就是……因为\(loj\)可以用了，所以以后在\(loj\)上做了~~（信息学奥赛一本通官方网站真是太\(rubbish\)了……）~~
\(ps:\)由于前几天已经把第一部分第一章做完了，所以从第二章开始写

二分与三分

首先……二分是一种非常精妙的算法，这个东西要用的话，必须要满足单调性，使用二分一次的时间约为\(O(n\log n)\)

LOJ #10011. 「一本通 1.2 例 1」愤怒的牛

题目链接

题意：数轴上有\(n\)个点，要求取出其中\(m\)个点，使得相邻的点的最小距离最大，输出这个距离

很明显的一道二分答案的题目，输入每个点的位置后从小到大排序，直接二分枚举答案，\(check\)函数比较简单，如下

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;

const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

int n, c;
int a[A];

bool check(int d) {
	int cow = 1;
	int now = a[1] + d;
	for(int i = 2; i <= n; i++) {
		if(a[i] < now) continue;
		cow++;
		now = a[i] + d;
	}
	return cow >= c;
}

signed main() {
	n = read(), c = read();
	for(int i = 1; i <= n; i++) a[i] = read();
	stable_sort(a + 1, a + n + 1);
	int l = 0, r = a[n] - a[1];
	while(l <= r) {
		int mid = (l + r) >> 1;
		if(check(mid)) l = mid + 1;
		else r = mid - 1;
	}
	cout << r << '\n'; return 0;
}

LOJ #10012. 「一本通 1.2 例 2」Best Cow Fences

题目链接

题意：给出\(N\)个数字\(A_i\)，求一段长度大于等于\(K\)的连续的子序列，其平均值最大

这道题属于实数二分，求平均值最大可以转化为每次\(check\)减去平均值看答案是否大于\(0\)即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ex 1e-5
using namespace std;

const int M = 10000100;
const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

double l = 100000.0, r = -10000.0, mid;
int n, L;
double a[M],  sum[M];

inline bool check(double ans) {
	double sum1, sum2;
	sum1 = sum[L-1] - (L-1) * ans;
	for(int i = L; i <= n; i++) {
		sum2 = sum[i] - sum[i - L] - L * ans;
		sum1 = sum1 + a[i] - ans;
		sum1 = max(sum1, sum2);
		if(sum1 > -ex) return true;
	}
	return false;
}

int main() {
	n = read(),  L = read();
	for(int i = 1; i <= n; i++) {
		scanf("%lf",  &a[i]);
		sum[i] = sum[i-1] + a[i];
		l = min(l, a[i]),  r = max(r, a[i]);
	}
	while(r - l > ex) {
		mid = (l + r)/2;
		if(check(mid)) l = mid;
		else r = mid;
	}
	int ans = (int)(r * 1000);
	return cout << ans << '\n',  0;
}

LOJ #10013. 「一本通 1.2 例 3」曲线

题目链接

题意：求一个函数\(F(x)\)在\([0,1000]\)上的最小值

这题就要用到三分了，三分是个啥?? here

由于函数\(S\)是开口向上的二次函数(\(a=0\)时为一次函数)，所以它是一个先单调递减，再单调递增的下凸函数，或者是一个一次的单调函数，\(F(x)=max(S(x))\)也是满足单调性的，所以就可以用到三分求极值（这里是最小值）

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

int test, n;
double a[A], b[A], c[A];
double x, maxx=0, L, r, Lmid, rmid;

double cal(double x) {
	double maxx = -0x7fffffff;
	for(int i = 1; i <= n; i++) maxx = max(maxx, a[i] * x * x + b[i] * x + c[i]);
	return maxx;
}

int main() {
	int T = read();
	for(int test = 1; test <= T; test++) {
		cin >> n;
		for(int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];
		L=0, r=1000;
		while(L + 1e-11 < r) {
			Lmid = L + (r - L) / 3;
			rmid = r - (r - L) / 3;
			if(cal(Lmid) <= cal(rmid)) r = rmid;
			else L = Lmid;
		}
		printf("%.4lf\n", cal(L));
	}
	return 0;
}

LOJ #10014. 「一本通 1.2 练习 1」数列分段 II

题目链接

题意：给定一个长度为\(n\)的正整数数列，现要将其分成\(m\)段，并要求每段连续，且每段和的最大值最小，求出这个最大的最小值

简单二分，在洛谷上也有 双倍经验啦

需要注意的是二分的下界\(l\)应为数列中的最大值，不然可能会\(wa\)

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;

const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

int ans = 0, sum = 0, maxn = -inf, n, m, a[A];

bool check(int x) {
	int now = 0, res = 1;
	for(int i = 1; i <= n; i++) {
		if(now + a[i] <= x) {
			now += a[i];
		} else {
			res++;
			now = a[i];
		}
	}
	return res <= m;
}

signed main() {
	n = read(), m = read();
	for(int i = 1; i <= n; i++) a[i] = read(), sum += a[i], maxn = max(maxn, a[i]);
	int l = maxn, r = sum;
	while(l <= r) {
		int mid = (l + r) >> 1;
		if(check(mid)) r = mid - 1;
		else l = mid + 1;
	}
	cout << l << '\n';
	return 0;
}

LOJ #10015. 「一本通 1.2 练习 2」扩散

题目链接

并查集+二分，有点懵逼。。先鸽着，可以看这篇博客点我

任意两点的距离为两点间的曼哈顿距离

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int A = 5e2 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

struct node { int x, y; } a[A];
int n, d[A][A], fa[A];
int abs(int x) { return x > 0 ? x : -x; }

int juli(node n1, node n2) { 
	return abs(n1.x - n2.x) + abs(n1.y - n2.y); 
}

int find(int x) { 
	return x == fa[x] ? x : fa[x] = find(fa[x]);
}

bool check(int x) {
	for(int i = 1; i <= n; i++) fa[i] = i;
	for(int i = 1; i <= n; i++)
		for(int j = i + 1; j <= n; j++)
			if(d[i][j] <= x * 2) {
				int fx = find(i), fy = find(j);
				if(fx != fy) fa[fx] = fy;
			}
	int sum = 0;
	for(int i = 1; i <= n; i++) {
		if(fa[i] == i) sum++;
		if(sum == 2) return false;
	}
	return true;
}

int main() {
	n = read();
	for(int i = 1; i <= n; i++) a[i].x = read(), a[i].y = read();
	for(int i = 1; i <= n; i++)
		for(int j = 1; j < i; j++)
			d[i][j] = d[j][i] = juli(a[i], a[j]);
	int l = 0, r = 999999999, ans = 0;
	while(l <= r) {
		int mid = (l + r) >> 1;
		if(check(mid)) r = mid - 1, ans = mid;
		else l = mid + 1;
	}
	cout << ans << '\n'; return 0;
}

LOJ #10016. 「一本通 1.2 练习 3」灯泡

题目链接

这个题就是要求影子的最大长度，假设在墙上的长度为\(x\)，我们就可以用相似得到\(L\)的表达式

\[L=D\times \dfrac {h-x}{H-x} +x \]

变形一下，就能够得到

\[L= \dfrac {D\times (H-h)}{x-H} + (x - H) + H + D \]

然后就发现这是一个对勾函数，说明可以进行三分，写就完了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define eps 1e-5
using namespace std;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

double H, D, h;

double Get(double x) {
	return D * (h - x) / (H - x) + x;
}

int main() {
	int T = read();
	while(T--) {
		scanf("%lf%lf%lf", &H, &h, &D);
		double l = -0, r = h;
		while(r - l > eps) {
			double qwq = (r - l) / 3, lmid = l + qwq, rmid = r - qwq;
			if(Get(lmid) < Get(rmid)) l = lmid;
			else r = rmid; 
		}
		printf("%.3lf\n", Get(l));
	}
	return 0;
}

LOJ #10017. 「一本通 1.2 练习 4」传送带

题目链接

神奇的三分套三分，我觉得题解讲的不错\(qwq\)

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const double eps = 1e-8;
double ax, ay, bx, by, cx, cy, dx, dy, p, q, r;

double dis(double x1, double y1, double x2, double y2) {
	double xdis = x1 - x2, ydis = y1 - y2;
	return sqrt(xdis * xdis + ydis * ydis);
}

double f(double x1, double y1, double x2, double y2) {
	return dis(x1, y1, x2, y2)/r + dis(x2, y2, dx, dy)/q;
}

double calc1(double x, double y) {
	double lx = cx, ly = cy, rx = dx, ry = dy;
	while(dis(lx, ly, rx, ry) > eps) {
		double kl = (rx - lx) / 3.0, ky = (ry - ly) / 3.0;
		double lmidx = lx + kl, rmidx = rx - kl, lmidy = ly + ky, rmidy = ry - ky;
		double ans1 = f(x, y, lmidx, lmidy), ans2 = f(x, y, rmidx, rmidy);
		if(ans2 - ans1 > eps) rx = rmidx, ry = rmidy;
		else lx = lmidx, ly = lmidy;
	}
	return f(x, y, lx, ly);
}

double calc() {
	double lx = ax, ly = ay, rx = bx, ry = by;
	while(dis(lx, ly, rx, ry) > eps) {
		double kl = (rx - lx) / 3.0, ky = (ry - ly) / 3.0;
		double lmidx = lx + kl, rmidx = rx - kl, lmidy = ly + ky, rmidy = ry - ky;
		double ans1 = calc1(lmidx, lmidy) + dis(ax, ay, lmidx, lmidy)/p, ans2 = calc1(rmidx, rmidy) + dis(ax, ay, rmidx, rmidy)/p;
		if(ans2 - ans1 > eps) rx = rmidx, ry = rmidy;
		else lx = lmidx, ly = lmidy;
	}
	return calc1(lx, ly) + dis(ax, ay, lx, ly)/p;
}

int main() {
	scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &ax, &ay, & bx, &by, &cx, &cy, &dx, &dy);
	scanf("%lf%lf%lf", &p, &q, &r);
	printf("%.2lf", calc());
}

posted @ 2019-12-19 15:12 Loceaner 阅读(1666) 评论(3) 编辑收藏举报

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「笔记」二分与三分

二分与三分

LOJ #10011. 「一本通 1.2 例 1」愤怒的牛

LOJ #10012. 「一本通 1.2 例 2」Best Cow Fences

LOJ #10013. 「一本通 1.2 例 3」曲线

LOJ #10014. 「一本通 1.2 练习 1」数列分段 II

LOJ #10015. 「一本通 1.2 练习 2」扩散

LOJ #10016. 「一本通 1.2 练习 3」灯泡

LOJ #10017. 「一本通 1.2 练习 4」传送带

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