poj 2373 Dividing the Path
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 2858 | Accepted: 1064 |
Description
To make installation easier, each sprinkler head must be installed along the ridge of the hill (which we can think of as a one-dimensional number line of length L (1 <= L <= 1,000,000); L is even).
Each sprinkler waters the ground along the ridge for some distance in both directions. Each spray radius is an integer in the range A..B (1 <= A <= B <= 1000). Farmer John needs to water the entire ridge in a manner that covers each location on the ridge by exactly one sprinkler head. Furthermore, FJ will not water past the end of the ridge in either direction.
Each of Farmer John's N (1 <= N <= 1000) cows has a range of clover that she particularly likes (these ranges might overlap). The ranges are defined by a closed interval (S,E). Each of the cow's preferred ranges must be watered by a single sprinkler, which might or might not spray beyond the given range.
Find the minimum number of sprinklers required to water the entire ridge without overlap.
Input
* Line 2: Two space-separated integers: A and B
* Lines 3..N+2: Each line contains two integers, S and E (0 <= S < E <= L) specifying the start end location respectively of a range preferred by some cow. Locations are given as distance from the start of the ridge and so are in the range 0..L.
Output
Sample Input
2 8
1 2
6 7
3 6
Sample Output
3
Hint
Two cows along a ridge of length 8. Sprinkler heads are available in integer spray radii in the range 1..2 (i.e., 1 or 2). One cow likes the range 3-6, and the other likes the range 6-7.
OUTPUT DETAILS:
Three sprinklers are required: one at 1 with spray distance 1, and one at 4 with spray distance 2, and one at 7 with spray distance 1. The second sprinkler waters all the clover of the range like by the second cow (3-6). The last sprinkler waters all the clover of the range liked by the first cow (6-7). Here's a diagram:
|-----c2----|-c1| cows' preferred ranges
|---1---|-------2-------|---3---| sprinklers
+---+---+---+---+---+---+---+---+
0 1 2 3 4 5 6 7 8
The sprinklers are not considered to be overlapping at 2 and 6.
Source
// dp[i]=min{dp[j]}+1 i-2*B=<j<=i-2*A
// 单调队列优化 求区间 dp[i-2*B] ~ dp[i-2*A] 最小值
#include <iostream>
#include <algorithm>
#include <queue>
#include <math.h>
#include <stdio.h>
#include <string.h>
using namespace std;
#define MOD 1000000007
#define maxn 1000010
int dp[maxn];
bool ok[maxn];
struct node{
int l,r; // 左右 下标 数值
node(){}
node(int a,int b){l=a;r=b;}
}in[],q[maxn];
int flag;
int N,L;
int A,B;
void solve(){
int i,j;
int head=,tail=,tp;
for(i=;i<=N;i++)
for(j=in[i].l+;j<in[i].r;j++)
ok[j]=;
for(i=*A;i<=L;i+=){
tp=i-*A;
if(ok[tp]){
while(head<tail&&dp[tp]<=q[tail-].r)
tail--;
if(ok[tp])
q[tail++]=node(tp,dp[tp]);
}
tp=i-*B;
while(head<tail&&q[head].l<tp)
head++; if(ok[i]&&head<tail) dp[i]=q[head].r+;//因为这里 所以最后结果可能 >=MOD wa的我好难过
}
}
int main()
{ dp[]=;
while(scanf("%d %d",&N,&L)!=EOF){
scanf("%d %d",&A,&B);
int i;
flag=true;
for(i=;i<=N;i++){
scanf("%d %d",&in[i].l,&in[i].r);
if(in[i].r-in[i].l>*B)
flag=false;
}
ok[]=;
if(flag){
for(i=;i<=L;i++)
ok[i]=,dp[i]=MOD;
solve();
}
if(!flag||dp[L]>=MOD) // 就是因为这里没有写 成 >= 而写成 == wa的我好难过 不过还好 OMs
printf("-1\n");
else
printf("%d\n",dp[L]);
} return ;
}
poj 2373 Dividing the Path的更多相关文章
- POJ 2373 Dividing the Path(DP + 单调队列)
POJ 2373 Dividing the Path 描述 农夫约翰的牛发现,在他的田里沿着山脊生长的三叶草是特别好的.为了给三叶草浇水,农夫约翰在山脊上安装了喷水器. 为了使安装更容易,每个喷头必须 ...
- POJ 2373 Dividing the Path (单调队列优化DP)题解
思路: 设dp[i]为覆盖i所用的最小数量,那么dp[i] = min(dp[k] + 1),其中i - 2b <= k <= i -2a,所以可以手动开一个单调递增的队列,队首元素就是k ...
- 【POJ】2373 Dividing the Path(单调队列优化dp)
题目 传送门:QWQ 分析 听说是水题,但还是没想出来. $ dp[i] $为$ [1,i] $的需要的喷头数量. 那么$ dp[i]=min(dp[j])+1 $其中$ j<i $ 这是个$ ...
- [POJ 2373][BZOJ 1986] Dividing the Path
Link: POJ 2373 传送门 Solution: 一开始想错方向的一道简单$dp$,不应该啊…… 我一开始的想法是以$cows' ranges$的节点为状态来$dp$ 但明显一个灌溉的区间的两 ...
- poj 3764 The xor-longest Path(字典树)
题目链接:poj 3764 The xor-longest Path 题目大意:给定一棵树,每条边上有一个权值.找出一条路径,使得路径上权值的亦或和最大. 解题思路:dfs一遍,预处理出每一个节点到根 ...
- poj2373 Dividing the Path
Dividing the Path Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5060 Accepted: 1782 ...
- Dividing the Path POJ - 2373(单调队列优化dp)
给出一个n长度的区间,然后有一些小区间只能被喷水一次,其他区间可以喷水多次,然后问你要把这个区间覆盖起来最小需要多少喷头,喷头的半径是[a, b]. 对于每个只能覆盖一次的区间,我们可以把他中间的部分 ...
- Dividing the Path POJ - 2373 dp
题意:你有无数个长度可变的区间d 满足 2a<=d<=2b且为偶数. 现在要你用这些区间填满一条长为L(L<1e6且保证是偶数)的长线段. 满足以下要求: 1.可变区间之间不能有 ...
- [USACO2004][poj2373]Dividing the Path(DP+单调队列)
http://poj.org/problem?id=2373 题意:一条直线分割成N(<=25000)块田,有一群奶牛会在其固定区域吃草,每1把雨伞可以遮住向左右延伸各A到B的区域,一只奶牛吃草 ...
随机推荐
- 编译为 Release 与 Debug 的区别
class Program { static void Main(string[] args) { DoWork(); } static void DoWork() { new Person().Ru ...
- [转载]c# OpenFileDialog
string resultFile = ""; OpenFileDialog openFileDialog1 = new OpenFileDialog(); ...
- uva 11825
刘书上例题 关于集合的动态规划 #include <cstdio> #include <cstdlib> #include <cmath> #include &l ...
- uva 1056
floyd 算法 用了stl 的map 存名字的时候比较方便 #include <cstdio> #include <cstdlib> #include <cmath&g ...
- android 解析XML方式(二)
上一节中,我们使用DOM方式解析xml文档,该方式比较符合我们日常思维方式,容易上手,但是它直接把文档调入内存中,比较耗内存.在这里我们可以用另外一种方式解析xml,这个就是SAX方式. SAX即是: ...
- 同一机器 部署 两个 jboss
当jboss和oracle在同一机器上时,通常oracle占用8080端口,这时只需要去修改\deploy\jbossweb-tomcat50.sar\server.xml中.当在同一台机器上运行两个 ...
- 微软发布WP SDK8.0 新增语音、应用内支付等原生API
http://www.csdn.net/article/2012-10-31/2811338-windows-phone-8-sdk 京时间10月30日,微软在旧金山举行新一代手机操作系统Window ...
- JS 变量或参数是否有值的判断
var node; …… 判断 node 是否有值,是否为 undefine,是否 null,直接使用两个!!,否定之否定: if (!!node){ .... }else{ .... } 这个条件判 ...
- 如何防止通过IP地址访问Tomcat管理页面
方法:建议修改webapps下面的原始文件夹的名称,比如加一个后缀: 当需要用管理页面的时候,可以将含有manager的文件夹的后缀去掉即可 manager和host-manager共2个文件夹
- Git PHP提交
做了个小的DEMO,可以查看: https://github.com/feixiang/webgit.git 这几天一直在郁闷的事情. Git在shell里面执行得好好的,apache运行用户也改成了 ...