Winter(bfs&&dfs)
| 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
- 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 ...
- 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!!!-------------------- ...
- Collect More Jewels(hdu1044)(BFS+DFS)
Collect More Jewels Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Othe ...
- Cleaning Robot (bfs+dfs)
Cleaning Robot (bfs+dfs) Here, we want to solve path planning for a mobile robot cleaning a rectangu ...
- 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 ...
- 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 ...
- 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 ...
- 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. ...
- 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
- JavaSE复习日记 : 抽象类
/* * 抽象类 * 抽象: * 面向对象的三大核心思想; * 封装: 封装,ppp是封装的一部分 * 继承; * 多态: 父类的引用指向子类的对象 * 引用: 是指一个引用型变量 * 有哪些变量? ...
- Cglib学习站点(转)
1.CGlib简单介绍,访问地址:http://www.blogjava.net/stone2083/archive/2008/03/16 /186615.html,从简单的示例到不同业务场景的变化, ...
- codeforces 455C 并查集
传送门 给n个点, 初始有m条边, q个操作. 每个操作有两种, 1是询问点x所在的连通块内的最长路径, 就是树的直径. 2是将x, y所在的两个连通块连接起来,并且要合并之后的树的直径最小,如果属于 ...
- while和for可以相互转换例子
//while和for循环可以相互转换,以下为简单格式: ;;) A; 等价于 : ) { A; ; } /* Name:while和for可以相互转换例子 Copyright: By.不懂网络 Au ...
- 在Ubuntu 11.10工具栏上用数字显示网速、CPU负荷和内存占用量『译』
基本上照抄了<How To Display Network Upload / Download Speed On The Panel In Ubuntu 11.04>,只不过我的实践环境是 ...
- 查看ORACLE 数据库及表信息
-- 查看ORACLE 数据库中本用户下的所有表 SELECT table_name FROM user_tables; -- 查看ORACLE 数据库中所有用户下的所有表 select user,t ...
- JAVA并发,CountDownLatch使用
该文章转自:http://www.itzhai.com/the-introduction-and-use-of-a-countdownlatch.html CountDownLatch 1.类介绍 一 ...
- android网络通讯数据封装之 json
Demo程序包括客户端和服务端 客户端按json数据格式封装数据传至服务端. 服务端为简单的servlet程序,负责接收客户端传到json数据,然后按原数据返回客户端. 实例代码如下: public ...
- oracle 11g RAC Grid Infrastructure
grid infrastructure 软件介质下载: http://www.oracle.com/technetwork/database/database-technologies/cluster ...
- PHPRPC for PHP
14的路 PHPRPC for PHP PHPRPC 是一个轻型的.安全的.跨网际的.跨语言的.跨平台的.跨环境的.跨域的.支持复杂对象传输的.支持引用参数传递的.支持内容输出重定向的.支持分级错误处 ...