Minimal coverage  The Problem Given several segments of line (int the X axis) with coordinates [Li,Ri]. You are to choose the minimal amount of them, such they would completely cover the segment [0,M]. The Input The first line is the number of test…
Given several segments of line (int the X axis) with coordinates [Li, Ri]. You are to choose the minimal amount of them, such they would completely cover the segment [0, M].InputThe first line is the number of test cases, followed by a blank line.Eac…
10020 Given several segments of line (int the X axis) with coordinates [Li, Ri]. You are to choose the minimalamount of them, such they would completely cover the segment [0, M].InputThe first line is the number of test cases, followed by a blank lin…
Minimal coverage The Problem Given several segments of line (int the X axis) with coordinates [Li,Ri]. You are to choose the minimal amount of them, such they would completely cover the segment [0,M]. The Input The first line is the number of test ca…
可以说是区间覆盖问题的例题... Note: 区间包含+排序扫描: 要求覆盖区间[s, t]; 1.把各区间按照Left从小到大排序,如果区间1的起点大于s,则无解(因为其他区间的左起点更大):否则选择起点在s的最长区间; 2.选择区间[li, ri]后,新的起点应更新为ri,并且忽略所有区间在ri之前的部分:  Minimal coverage  The Problem Given several segments of line (int the X axis) with coordinat…
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=961 贪心,排序,对左端点贪,找最大右端点. #include <cstdio> #include <cstring> #include <algorithm> #define maxn 6000000 using namespace std; int t,…
n sprinklers are installed in a horizontal strip of grass l meters long and w meters wide. Each sprinkler is installed at the horizontal center line of the strip. For each sprinkler we are given its position as the distance from the left end of the c…
链接: http://acm.timus.ru/problem.aspx?space=1&num=1303 按照贪心的思想,每次找到覆盖要求区间左端点时,右端点最大的线段,然后把要求覆盖的区间改为这个右端点到M这个区间.依次类推下去,这样的话就只需要扫一遍就可以找去来. 要做的预备工作就是将线段按照左端点的升序排序就可以了. 它的时间复杂度就是O(n) 代码一直WA,望大神指教 #include<iostream> #include<stdio.h> #include<…
题目: 给定平面上n(n≤105)个点和一个值D,要求在x轴上选出尽量少的点,使得对于给定的每个点,都有一个选出的点离它的欧几里得距离不超过D. 思路: 先自己造区间,然后贪心选点就可以了.之前做过一到类似的题目还是没有一眼看出来. 区间的造法,就是以给出的点为圆心,以D为半径画圆,这个圆与x轴的相交的两个点就是我们造的区间的左右两个端点. 然后以右端点从小到大排序贪心.(自己举了一个例子发现左端点要比右端点多) 代码: #include <bits/stdc++.h> #define inf…
题目大意:先确定一个M, 然后输入多组线段的左端和右端的端点坐标,然后让你求出来在所给的线段中能够 把[0, M] 区域完全覆盖完的最少需要的线段数,并输出这些线段的左右端点坐标. 思路分析: 线段区间的起点是0,那么找出所有区间起点小于0中的最合适的区间. 因为需要尽量少的区间,所以选择右端点更大的区间,它包含所选线段更大. 如果在所有区间中找到了解,且右端点小于M,则把找到的区间的右端点定为新的线段区间的起点. #include <iostream> #include <stdio.…