【二分】Caravan Robbers

Caravan Robbers

Long long ago in a far far away land there were two great cities and The Great Caravan Road between them. Many robber gangs “worked” on that road.
By an old custom the i-th band robbed all merchants that dared to travel between ai and bi miles of The Great Caravan Road. The custom was old, but a clever one, as there were no two distinct i and j such that ai ≤ aj and bj ≤ bi. Still when intervals controlled by two gangs intersected, bloody fights erupted occasionally. Gang leaders decided to end those wars. They decided to assign each gang a new interval such that all new intervals do not intersect (to avoid bloodshed), for each gang their new interval is subinterval of the old one (to respect the old custom), and all new intervals are of equal length (to keep things fair).
You are hired to compute the maximal possible length of an interval that each gang would control after redistribution.
Input
The input will contain several test cases, each of them as described below.
The first line contains n (1 ≤ n ≤ 100000) — the number of gangs. Each of the next n lines contains
information about one of the gangs — two integer numbers ai and bi (0 ≤ ai < bi ≤ 1000000). Data
provided in the input file conforms to the conditions laid out in the problem statement.
Output
For each test case, write to the output on a line by itself.
Output the maximal possible length of an interval in miles as an irreducible fraction p/q.
Note for the sample:
In the above example, one possible set of new intervals that each gang would control after redistribution is given below.
• The first gang would control an interval between 7/2 = 3.5 and 12/2 = 6 miles which has length
of 5/2 and is a subinterval of its original (2, 6).
• The second gang would control an interval between 2/2 = 1 and 7/2 = 3.5 miles which has length
of 5/2 and is a subinterval of its original (1, 4).
• The third gang would control an interval between 16/2 = 8 and 21/2 = 10.5 miles which has
length of 5/2 and is a subinterval of its original (8, 12).
Sample Input

Sample Output

5/2

题目大意

给出一些区间，这些区间可能部分重叠，现请你求出一个最大的区间长度 lll，使得给出的所有区间都变成这个长度 lll 并且所有区间都不重叠。

思路

二分答案
一个关键点在于判断某个区间长度是否符合题意
因为这道题的答案是用分数表示的，但是不能去二分分数，因为用分数的话会很复杂而且分子分母可能很大很大，所以只能二分小数，然后把小数转成分数，所以另一个关键点在于如何把这个小数变成分数。

判断某个区间长度是否符合题意：

先排序区间，将起点从小到大排序，当起点一样时终点小的排在前面
设置一个起始指针，从头开始遍历每个区间，如果这个指针小于当前区间的起点的话，就把这个指针拨到区间的起点，如果起始指针加上待判断的区间长度大于当前区间的终点时，就返回假，表示不符合题意；否则起始指针就拨到加上了待判断区间长度的位置。直到所有区间都判断完了，如果都行，那么就返回真。

把小数变成分数：

二分答案出来了之后，我们需要把这个答案变成分数表示，即把这个小数表示为 p/qp/qp/q 的形式。
设这个二分答案为 ansansans，就有ans=p/qans=p/qans=p/q；如果枚举分母 qqq，以此确定分子 qqq，并使得最后得到的 p/qp/qp/q 最接近 ansansans，那么最后就确定了 ppp和 qqq。
这里记录一种很“巧妙”的方法。

代码如下

#include <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <deque>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define LL long long
#define UIT unsigned int
#define DB double
#define PRII pair<int, int>
#define MMST(a) memset(a, 0, sizeof(a))
#define FOR(i, beg, ed, s) for (int i = beg; i < ed; i += s)
#define FORE(i, beg, ed, s) for (int i = beg; i <= ed; i += s)
#define LOOP while (true)
#define in scanf
#define out printf
using namespace std;const DB EPS = 1e-10;vector<pair<int, int>> aa;
int n;//判断某个区间长度是否符合题意
bool check(DB x) {DB st = 0;FOR(i, 0, n, 1) {if (st < aa[i].first) st = aa[i].first;if (st + x <= aa[i].second) {st += x;} else {return false;}}return true;
}void solve() {int a = 0, b = 0;DB r = 0, l = 0, mid = 0;LL p = 0, q = 0, ux = 0;//多组数据while (~in("%d", &n)) {aa.clear();//不要忘了清空容器FOR(i, 0, n, 1) {in("%d%d", &a, &b);aa.push_back(make_pair(a, b));}sort(aa.begin(), aa.end());l = 0, r = INT_MAX;while (r - l > EPS) {mid = l + (r - l) / 2.0;if (check(mid)) {l = mid;} else {r = mid;}}//这里得到二分答案l，并将其转为分数p = 0, q = 1;FORE(vx, 1, n, 1) {ux = round(l * vx);if (fabs((DB)ux / vx - l) < fabs((DB)p / q - l)) {p = ux, q = vx;}}out("%lld/%lld\n", p, q);}
}int main(void) {solve();return 0;
}

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【二分】Caravan Robbers

Caravan Robbers

题目大意

思路

代码如下

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