Cleaning Shifts
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 4715   Accepted: 1590

Description

Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most obliging of farmers, has no choice but hire some of the cows to clean the barn.

Farmer John has N (1 <= N <= 10,000) cows who are willing to do some cleaning. Because dust falls continuously, the cows require that the farm be continuously cleaned during the workday, which runs from second number M to second number E during the day (0 <= M <= E <= 86,399). Note that the total number of seconds during which cleaning is to take place is E-M+1. During any given second M..E, at least one cow must be cleaning.

Each cow has submitted a job application indicating her willingness to work during a certain interval T1..T2 (where M <= T1 <= T2 <= E) for a certain salary of S (where 0 <= S <= 500,000). Note that a cow who indicated the interval 10..20 would work for 11 seconds, not 10. Farmer John must either accept or reject each individual application; he may NOT ask a cow to work only a fraction of the time it indicated and receive a corresponding fraction of the salary.

Find a schedule in which every second of the workday is covered by at least one cow and which minimizes the total salary that goes to the cows.

Input

Line 1: Three space-separated integers: N, M, and E.

Lines 2..N+1: Line i+1 describes cow i's schedule with three space-separated integers: T1, T2, and S.

Output

Line 1: a single integer that is either the minimum total salary to get the barn cleaned or else -1 if it is impossible to clean the barn.

Sample Input

3 0 4
0 2 3
3 4 2
0 0 1

Sample Output

5

Hint

Explanation of the sample:

FJ has three cows, and the barn needs to be cleaned from second 0 to second 4. The first cow is willing to work during seconds 0, 1, and 2 for a total salary of 3, etc.

Farmer John can hire the first two cows.

 
题意: n个区间 每个区间[ l , r ]有一个权值w ,  问覆盖区间[ m , e ]的最小权值是多少。
解析: 首先按照右端点从小到大排序,然后dp[i][j] 表示 考虑前i个区间,覆盖区间[m,j]的最小值。
考虑当前区间[ li , ri ]容易想到 dp[i][ri] 的最小值就是 dp[ri]=min(dp[ri],min(dp[ li -1, ri ])+wi) 我们可以线段树查找区间最小值进行优化,
当我们确定了dp[ri] 其实更新ri这一个点值就可以了 因为下一个区间[ Li+1,Ri+1 ] 询问最小值的时候,一定会询问到dp[ri],询问不到,说明 Li+1是大于 ri 的,覆盖的区间已经断开了。
AC代码
#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <algorithm>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("\n")
#define debug(a,b) cout<<a<<" "<<b<<" "<<endl
#define ffread(a) fastIO::read(a)
using namespace std;
typedef long long ll;
const int maxn = 1e5+;
const ll inf = 1e12;
const ll mod = ;
const double epx = 1e-;
const double pi = acos(-1.0);
//head-----------------------------------------------------------------
ll minn[maxn*];
void PushUp(int rt)
{
minn[rt]=min(minn[rt<<],minn[rt<<|]);
}
void update(int x,ll val,int l,int r,int rt)
{
if(l==x&&r==x)
{
minn[rt]=val;
return;
}
int mid=(l+r)>>;
if(x<=mid)
update(x,val,l,mid,rt<<);
else
update(x,val,mid+,r,rt<<|);
PushUp(rt);
}
ll query(int L,int R,int l,int r,int rt)
{
if(L<=l&&R>=r)
{
return minn[rt];
}
int mid=(l+r)>>;
ll ans=1e12;
if(L<=mid) ans=min(ans,query(L,R,l,mid,rt<<));
if(R>mid) ans=min(ans,query(L,R,mid+,r,rt<<|));
return ans;
}
struct node
{
ll l,r,val;
}qujian[maxn];
bool cmp(node a,node b)
{
return a.r<b.r;
}
int main()
{
ll n,m,e;
scanf("%lld%lld%lld",&n,&m,&e);
m+=,e+=;
for(int i=;i<n;i++)
{
scanf("%lld%lld%lld",&qujian[i].l,&qujian[i].r,&qujian[i].val);
qujian[i].l+=,qujian[i].r+=;
}
sort(qujian,qujian+n,cmp);
for(int i=;i<=1e5;i++)
update(i,inf,,1e5,);
update(m-,,,1e5,);
for(int i=;i<n;i++)
{
if(qujian[i].r>=m)
{
ll tempmin=query(qujian[i].l-,qujian[i].r,,1e5,);
if(tempmin==inf)
continue;
ll temp=tempmin+qujian[i].val;
ll before=query(qujian[i].r,qujian[i].r,,1e5,);
if(before>temp)
update(qujian[i].r,temp,,1e5,);
}
}
ll ans=query(e,1e5,,1e5,);
if(ans==inf)
printf("-1\n");
else
printf("%lld\n",ans);
}

POJ 3171 区间覆盖最小值&&线段树优化dp的更多相关文章

  1. POJ 1769 Minimizing maximizer (线段树优化dp)

    dp[i = 前i中sorter][j = 将min移动到j位置] = 最短的sorter序列. 对于sorteri只会更新它右边端点r的位置,因此可以把数组改成一维的,dp[r] = min(dp[ ...

  2. POJ 2376 Cleaning Shifts (线段树优化DP)

    题目大意:给你很多条线段,开头结尾是$[l,r]$,让你覆盖整个区间$[1,T]$,求最少的线段数 题目传送门 线段树优化$DP$裸题.. 先去掉所有能被其他线段包含的线段,这种线段一定不在最优解里 ...

  3. 洛谷$P2605\ [ZJOI2010]$基站选址 线段树优化$dp$

    正解:线段树优化$dp$ 解题报告: 传送门$QwQ$ 难受阿,,,本来想做考试题的,我还造了个精妙无比的题面,然后今天讲$dp$的时候被讲到了$kk$ 先考虑暴力$dp$?就设$f_{i,j}$表示 ...

  4. 2021.12.08 P1848 [USACO12OPEN]Bookshelf G(线段树优化DP)

    2021.12.08 P1848 [USACO12OPEN]Bookshelf G(线段树优化DP) https://www.luogu.com.cn/problem/P1848 题意: 当农夫约翰闲 ...

  5. [AGC011F] Train Service Planning [线段树优化dp+思维]

    思路 模意义 这题真tm有意思 我上下楼梯了半天做出来的qwq 首先,考虑到每K分钟有一辆车,那么可以把所有的操作都放到模$K$意义下进行 这时,我们只需要考虑两边的两辆车就好了. 定义一些称呼: 上 ...

  6. 4.11 省选模拟赛 序列 二分 线段树优化dp set优化dp 缩点

    容易想到二分. 看到第一个条件容易想到缩点. 第二个条件自然是分段 然后让总和最小 容易想到dp. 缩点为先:我是采用了取了一个前缀最小值数组 二分+并查集缩点 当然也是可以直接采用 其他的奇奇怪怪的 ...

  7. 【bzoj3939】[Usaco2015 Feb]Cow Hopscotch 动态开点线段树优化dp

    题目描述 Just like humans enjoy playing the game of Hopscotch, Farmer John's cows have invented a varian ...

  8. D - The Bakery CodeForces - 834D 线段树优化dp···

    D - The Bakery CodeForces - 834D 这个题目好难啊,我理解了好久,都没有怎么理解好, 这种线段树优化dp,感觉还是很难的. 直接说思路吧,说不清楚就看代码吧. 这个题目转 ...

  9. Codeforces 1603D - Artistic Partition(莫反+线段树优化 dp)

    Codeforces 题面传送门 & 洛谷题面传送门 学 whk 时比较无聊开了道题做做发现是道神题( 介绍一种不太一样的做法,不观察出决策单调性也可以做. 首先一个很 trivial 的 o ...

随机推荐

  1. Leetcode5078. 负二进制数相加

    问题: 5078. 负二进制数相加 给出基数为 -2 的两个数 arr1 和 arr2,返回两数相加的结果. 数字以 数组形式 给出:数组由若干 0 和 1 组成,按最高有效位到最低有效位的顺序排列. ...

  2. nginx url rewrite break和last的区别

    break 将重写的URI作为一个新的URI,在本块中继续处理,将重写后 的地址在当前location块中处理,不会将新的URI转向到其他location块中 last,终止继续在本location块 ...

  3. vue 顶级组件

    快 有时候懒的把一些通用组件写到template里面去,而业务中又需要用到,比如表示loading状态这样组件. 如果是这样的组件,可以选择把组件手动初始化,让组件在整个app生命周期中始终保持活跃. ...

  4. 点击tr实现选择checkbox功能,点击checkobx的时候阻止冒泡事件, jquery给checkbox添加checked属性或去掉checked属性不能使checkobx改变状态

    给tr添加点击事件,使用find方法查找tr下的所有层级的元素,children只查找下一层级的元素,所以使用find.find的返回值为jquery对象,在这个项目中不知道为什么使用jquery给c ...

  5. ccf 201712-2 游戏(Python实现)

    一.原题 问题描述 试题编号: 201712-2 试题名称: 游戏 时间限制: 1.0s 内存限制: 256.0MB 问题描述: 问题描述 有n个小朋友围成一圈玩游戏,小朋友从1至n编号,2号小朋友坐 ...

  6. Yii2.0 的安装学习

    视频学习地址: 后盾网视频: http://www.houdunren.com/houdunren18_lesson_76?vid=7350 与<Yii框架>不得不说的故事—基础篇 htt ...

  7. HDU - 5884 Sort (二分答案+贪心)

    有n个数字,你需要把这n个数字合成一个数字,每次只能把k个数字合并成一个,花费为这k个数字的和. 给一个最大花费,问不超过这个最大花费的情况下,k的最小值. Sample Input 1 5 25 1 ...

  8. python for data analysis chapter1~2

    Q1:numpy与series的区别:index Tab补全(任意路径Tab) 内省(函数:?显示文档字符串,??显示源代码:结合通配符:np.* load *?) %load .py ctrl-c( ...

  9. bzoj3262陌上花开 三维数点 cdq+树状数组

    大早上的做了一道三维数点一道五位数点,神清气爽! 先给一维排序,变成一个奇怪的动态的二维数点(相当于有一个扫描面扫过去,导致一系列的加点和询问) 然后cdq分治,再变回静态,考虑前半段对后半段的影响 ...

  10. NuGet安装本地包命令行

    尝试安装本地的NuGet包. 键入 "get-help NuGet" 可查看所有可用的 NuGet 命令. install-package Polly.Net40Async-Sig ...