题目大意.N个区间覆盖[T1,T2]及相应的代价S,求从区间M到E的所有覆盖的最小代价是多少. (1 <= N <= 10,000).(0 <= M <= E <= 86,399). 思路是DP,首先将每一个区间依照T2从小到大排序,设dp(k)为从m覆盖到k所需最小代价,则有 dp(T2[i]) = min(dp(T2[i]), {dp(j) + Si,  T1[i] - 1<=j <= T2[i]}),对于 {dp(j) + Si,  T1[i] - 1<…

/* http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?pid=1019&ojid=1&cid=10 题目: 给定一个时间T和N个时间区间,求最少需要多少个区间覆盖总区间[1,T],无法覆盖区域[1,T]时输出-1. 例如T=10,有3个区间[1,7],[3,6],[8,10],则最少需要两个区间来覆盖,选择区间1和区间3. 解题思路: 使用贪心法.首先将区间按开始时间从小到大排序,开始时间相等按结束时间从小到大排序. 1 如…

Cleaning Shifts Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 40751   Accepted: 9871 Description Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one co…

https://vjudge.net/problem/POJ-3171.(有价值的区间全覆盖问题) (lyd例题)朴素DP很好想,$f[i]$表示将右端点从小到大排序后从$L$(要求覆盖的大区间)到第$i$个区间的右端点都覆盖完成时最小化费.无解则为INF.然后利用排序.右端点单调性,每次枚举前面dp过的右端点落在现在左端点及其右侧(或相邻)的,去最小dp值加上现在费用.然后$O(n^2)$就可以水过啦..因为常数特别小,数据也才1e4嘛..优化的话因为每次查的都是一段区间里的右端点最值,所以线…

Cleaning Shifts Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4422   Accepted: 1482 Description Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer…

题目:http://poj.org/problem?id=3171 题意:给你n个区间[a,b],每个区间都有一个费用c,要你用最小的费用覆盖区间[M,E] 分析:经典的区间覆盖问题,百度可以搜到这个专题. 线段覆盖问题一般考虑贪心和DP,但是每个区间又有了一个费用c,本渣觉得贪心貌似不太行(不知道神犇门有木有贪心解法),然后便考虑DP 设f[i]表示覆盖[M,i]的最小总费用 那么有f[t[i].a]=min(f[j]+t[i].c) t[i].a-1<=j<=t[i].b-1 那么ans=…

<pre name="code" class="html"> Cleaning Shifts Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14425   Accepted: 3700 Description Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleani…

传送门 题目大意 有一个大区间和n个小区间,每个小区间都有一个代价,求最少付出多少代价可以使得小区间完全覆盖大区间. 分析为了方便起见我们先将s变为2,其它的位置都对应更改以便后期处理.我们考虑以t1为第一关键字,t2为第二关键字将所有奶牛排序.用dp[i][j]表示考虑到第i只牛,覆盖到点j最少需要多少钱.我们可以将i这一维去掉,则dp[j]=min{dp[j],dp[j'](t1-1<=j'<=t2-1)}.然后进行线段树优化就可以了. 代码 #include<iostream>…

https://vjudge.net/problem/POJ-2376 题意理解错了!!真是要仔细看题啊!! 看了poj的discuss才发现,如果前一头牛截止到3,那么下一头牛可以从4开始!!! #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #include<se…

题意 POJ2376 Cleaning Shifts 0x50「动态规划」例题 http://bailian.openjudge.cn/practice/2376 总时间限制: 1000ms 内存限制: 65536kB 描述 Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one c…