1084 - Winter
Time Limit: 2 second(s) Memory Limit: 32 MB

Winter is coming. In a land far away, N men are spending the nights in a valley in a largest field. The valley is so narrow that it can be considered to be a straight line running east-to-west.

Although standing in the valley does shield them from the wind, the group still shivers during the cold nights. They, like anyone else, would like to gather together for warmth.

Near the end of each day, each man i finds himself somewhere in the valley at a unique location Li. The men want to gather into groups of three or more persons since two persons just aren't warm enough. They want to be in groups before sunset, so the distance K each man can walk to form a group is limited. Determine the smallest number of groups the men can form.

Input

Input starts with an integer T (≤ 15), denoting the number of test cases.

Each case starts with two integers N (1 ≤ N ≤ 105) and K (1 ≤ K ≤ 106). Each of the next N line contains an integer Li (1 ≤ Li ≤ 108).

Output

For each case, print the case number and smallest number of groups the men can gather into. If there is no way for all the men to gather into groups of at least size three, output -1.

Sample Input

Output for Sample Input

2

6 10

2

10

15

13

28

9

3 1

1 10 20

Case 1: 2

Case 2: -1

Note

Dataset is huge, use faster I/O methods.

题意:N个人要在日落前都组成>=3的组,最多走距离k;问组成最少的组数;不能输出-1;

思路:本来自己想着贪心的,最后发现自己想错了,因为聚集点是不确定的,而且要的是最小的组,看了巨巨的有dfs还有bfs的;于是就借鉴了巨巨的思路;感觉应该是优先队列的,巨巨竟然不是优先队列也对了,膜拜;

bfs:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define T_T while(T--)
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=1e5+100;
int a[MAXN];
int N,K;
int vis[MAXN];
/*struct Node{
int l,r;
friend bool operator < (Node a,Node b){
if(a.l!=b.l)return a.l<b.l;
else return a.r<b.r;
}
};
Node d[MAXN];*/
struct Node{
int pos,gg;
friend bool operator < (Node a,Node b){
return a.gg>b.gg;
}
};
priority_queue<Node>dl;
int bfs(){
Node p,q;
mem(vis,0);
vis[0]=1;
while(!dl.empty())dl.pop();
p.pos=0;p.gg=0;
dl.push(p);
while(!dl.empty()){
p=dl.top();dl.pop();
int temp=p.pos;
if(temp>=N)return p.gg;
while(temp<N&&a[temp]-a[p.pos]<=2*K)temp++;
//PI(temp);puts("");
if(temp-p.pos>=3){
q.pos=temp;q.gg=p.gg+1;
if(!vis[q.pos])
dl.push(q);
vis[q.pos]=1;
}
if(temp-p.pos>=4){
q.pos=temp-1;q.gg=p.gg+1;
if(!vis[q.pos])
dl.push(q);
vis[q.pos]=1;
}
if(temp-p.pos>=5){
q.pos=temp-2;q.gg=p.gg+1;
if(!vis[q.pos])
dl.push(q);
vis[q.pos]=1;
}
}
return -1;
}
int main(){
int T,kase=0;
SI(T);
T_T{
SI(N);SI(K);
for(int i=0;i<N;i++)SI(a[i]);
sort(a,a+N);
//for(int i=0;i<N;i++)PI(a[i]),P_;puts("");
/*int l,r;
for(int i=0;i<N;i++){
d[i].l=lower_bound(a,a+N,a[i]-K)-a;
d[i].r=upper_bound(a,a+N,a[i]+K)-a-1;
}
sort(d,d+N);*/
printf("Case %d: %d\n",++kase,bfs());
}
return 0;
}

  dfs:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define T_T while(T--)
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=1e5+100;
int a[MAXN];
int N,K;
int vis[MAXN];
/*struct Node{
int pos,gg;
friend bool operator < (Node a,Node b){
return a.gg>b.gg;
}
};
priority_queue<Node>dl;
int bfs(){
Node p,q;
mem(vis,0);
vis[0]=1;
while(!dl.empty())dl.pop();
p.pos=0;p.gg=0;
dl.push(p);
while(!dl.empty()){
p=dl.top();dl.pop();
int temp=p.pos;
if(temp>=N)return p.gg;
while(temp<N&&a[temp]-a[p.pos]<=2*K)temp++;
//PI(temp);puts("");
if(temp-p.pos>=3){
q.pos=temp;q.gg=p.gg+1;
if(!vis[q.pos])
dl.push(q);
vis[q.pos]=1;
}
if(temp-p.pos>=4){
q.pos=temp-1;q.gg=p.gg+1;
if(!vis[q.pos])
dl.push(q);
vis[q.pos]=1;
}
if(temp-p.pos>=5){
q.pos=temp-2;q.gg=p.gg+1;
if(!vis[q.pos])
dl.push(q);
vis[q.pos]=1;
}
}
return -1;
}
int main(){
int T,kase=0;
SI(T);
T_T{
SI(N);SI(K);
for(int i=0;i<N;i++)SI(a[i]);
sort(a,a+N);
//for(int i=0;i<N;i++)PI(a[i]),P_;puts("");
printf("Case %d: %d\n",++kase,bfs());
}
return 0;
}*/
int ans,flot;
void dfs(int pos,int gg){
int temp=pos;
if(flot)return;
if(pos>=N){
flot=1;
ans=gg;
return;
}
while(temp<N&&a[temp]-a[pos]<=2*K)temp++;
if(!vis[temp])if(temp-pos>=3)vis[temp]=1,dfs(temp,gg+1);
if(!vis[temp-1])if(temp-pos>=4)vis[temp-1]=1,dfs(temp-1,gg+1);
if(!vis[temp-2])if(temp-pos>=5)vis[temp-2]=1,dfs(temp-2,gg+1);
}
int main(){
int T,kase=0;
SI(T);
T_T{
SI(N);SI(K);
for(int i=0;i<N;i++)SI(a[i]);
sort(a,a+N);
//for(int i=0;i<N;i++)PI(a[i]),P_;puts("");
flot=0;
mem(vis,0);
vis[0]=1;
dfs(0,0);
if(!flot)ans=-1;
printf("Case %d: %d\n",++kase,ans);
}
return 0;
}

  

Winter(bfs&&dfs)的更多相关文章

  1. POJ 2227 The Wedding Juicer (优先级队列+bfs+dfs)

    思路描述来自:http://hi.baidu.com/perfectcai_/item/701f2efa460cedcb0dd1c820也可以参考黑书P89的积水. 题意:Farmer John有一个 ...

  2. 邻结矩阵的建立和 BFS,DFS;;

    邻结矩阵比较简单,, 它的BFS,DFS, 两种遍历也比较简单,一个用队列, 一个用数组即可!!!但是邻接矩阵极其浪费空间,尤其是当它是一个稀疏矩阵的时候!!!-------------------- ...

  3. Collect More Jewels(hdu1044)(BFS+DFS)

    Collect More Jewels Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Othe ...

  4. Cleaning Robot (bfs+dfs)

    Cleaning Robot (bfs+dfs) Here, we want to solve path planning for a mobile robot cleaning a rectangu ...

  5. LeetCode:BFS/DFS

    BFS/DFS 在树专题和回溯算法中其实已经涉及到了BFS和DFS算法,这里单独提出再进一步学习一下 BFS 广度优先遍历 Breadth-First-Search 这部分的内容也主要是学习了labu ...

  6. 图的基本遍历算法的实现(BFS & DFS)复习

    #include <stdio.h> #define INF 32767 typedef struct MGraph{ ]; ][]; int ver_num, edge_num; }MG ...

  7. BFS/DFS算法介绍与实现(转)

    广度优先搜索(Breadth-First-Search)和深度优先搜索(Deep-First-Search)是搜索策略中最经常用到的两种方法,特别常用于图的搜索.其中有很多的算法都用到了这两种思想,比 ...

  8. NOIP2010引水入城[BFS DFS 贪心]

    题目描述 在一个遥远的国度,一侧是风景秀美的湖泊,另一侧则是漫无边际的沙漠.该国的行政区划十分特殊,刚好构成一个N 行M 列的矩形,如上图所示,其中每个格子都代表一座城市,每座城市都有一个海拔高度. ...

  9. HDU 1044 Collect More Jewels(BFS+DFS)

    Collect More Jewels Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Othe ...

随机推荐

  1. realloc,c语言

    realloc #include <stdlib.h> main() { char* ptr=NULL; char* ptr2=NULL; ptr = malloc(); printf(& ...

  2. Struts学习之流程汇总

    struts2 架构图如下图所示: 依照上图,我们可以看出一个请求在struts的处理大概有如下步骤: 1.客户端初始化一个指向Servlet容器(例如Tomcat)的请求: 2.这个请求经过一系列的 ...

  3. jQuery实时获取checkbox状态问题

    在最近的项目开发中,使用jQuery操作checkbox时,发现一个问题. Html代码如下: <body> <div> <inputtype="checkbo ...

  4. Maven手动创建多模块项目

    Maven手动创建多模块项目 我要创建的项目名称是:unicorn,项目包含两个模块,分别是unicorn-core和unicorn-web.包的路径是com.goldpalm.tour. 项目创建流 ...

  5. 算法学习笔记(LeetCode OJ)

    ================================== LeetCode的一些算法题,都是自己做的,欢迎提出改进~~ LeetCode:http://oj.leetcode.com == ...

  6. [LeetCode]题解(python):142-Linked List Cycle II

    题目来源: https://leetcode.com/problems/linked-list-cycle-ii/ 题意分析: 给定一个链表,如果链表有环,返回环的起始位置,否则返回NULL.要求常量 ...

  7. MYSQL ERROR 1045 (28000): Access denied for user 'neeky'@'Nee' (using password: YES)

    情况: mysql -h 192.168.1.7 -u neeky -p 本来这样就可以连接上mysql服务的了, 可是它会报这个错“ERROR 1045 (28000): Access denied ...

  8. HUB主要芯片方案

    HUB主要品牌:慧荣.擎泰.联盛  安国.创惟 创惟GL850G简介:拥有低耗电.温度低及接脚数减少等产品特性.它支援4个下游连接埠,采用48 pin LQFP封装,可完全支援USB 2.0/1.1规 ...

  9. 视频主观质量评价工具:MSU Perceptual Video Quality tool

    MSU Perceptual Video Quality tool是莫斯科国立大学(Moscow State University)的Graphics and Media Lab制作的一款视频主观评价 ...

  10. Oracle中NVARCHAR2与VARCHAR2的差别

    NVARCHAR2在计算长度时和字符集相关的: 比如数据库是中文字符集时以长度10为例, 1.NVARCHAR2(10)是能够存进去10个汉字的.假设用来存英文也仅仅能存10个字符. 2.而VARCH ...