USACO 2018 February Contest, Bronze Hoofball

 Hoofball

时间限制: 1 Sec  内存限制: 128 MB
提交: 166  解决: 27
[提交][状态][讨论版][命题人:admin]

题目描述

In preparation for the upcoming hoofball tournament, Farmer John is drilling his N cows (conveniently numbered 1…N, where 1≤N≤100) in passing the ball. The cows are all standing along a very long line on one side of the barn, with cow ii standing xi units away from the barn (1≤xi≤1000). Each cow is standing at a distinct location.
At the begiNing of the drill, Farmer John will pass several balls to different cows. When cow i receives a ball, either from Farmer John or from another cow, she will pass the ball to the cow nearest her (and if multiple cows are the same distance from her, she will pass the ball to the cow farthest to the left among these). So that all cows get at least a little bit of practice passing, Farmer John wants to make sure that every cow will hold a ball at least once. Help him figure out the minimum number of balls he needs to distribute initially to ensure this can happen, assuming he hands the balls to an appropriate initial set of cows.

输入

The first line of input contains N. The second line contains N space-separated integers, where the ith integer is xi.

输出

Please output the minimum number of balls Farmer John must initially pass to the cows, so that every cow can hold a ball at least once.

样例输入

5

7 1 3 11 4

样例输出

2

提示

In the above example, Farmer John should pass a ball to the cow at x=1 and pass a ball to the cow at x=11. The cow at x=1 will pass her ball to the cow at x=3, after which this ball will oscillate between the cow at x=3 and the cow at x=4. The cow at x=11 will pass her ball to the cow at x=7, who will pass the ball to the cow at x=4, after which this ball will also cycle between the cow at x=3 and the cow at x=4. In this way, all cows will be passed a ball at least once (possibly by Farmer John, possibly by another cow).

It can be seen that there is no single cow to whom Farmer John could initially pass a ball so that every cow would eventually be passed a ball.

来源

USACO 2018 February Contest, Bronze 

【题意】

    扑朔迷离的英文题； 不， 是扑朔迷离的英语水平， 看不懂题意；；；；；；

    是这样的， 某个点传球， 传两边，  传给距离最小的， 其次，当距离相等时 传给左边；

    问最少需要几个球 让所有的人都曾经摸过球。

【思路】 

    想着类似于 波浪线这种， 有上升有下降， 取顶点，  对于 平台  取奇数的 ，  但是实现起来特别麻烦。

    用 环标记，   把 某个数 存左边还是右边  建立 关系，  构成 几个环， 然后DFS 搜一下环，

    最后 没有标记的点，就属于平台部分了，  然后/2  加上环的个数

【代码实现】

/*
* Date: 4/3/2018
* Tile: UPC 补题
* Category: DFS
* AU: siz
* WA: 1
* Attation: 标记后， 进行DFS 搜索，，最后扫描， 平台的情况
*/
#include <iostream>
#include <bits/stdc++.h>
typedef long long ll;
const int INF=0x3f3f3f3f;
const int MAXN=1e5+5;

using namespace std;

int head[MAXN],vis[MAXN];
int in[MAXN];
int cot,n,m;
struct node{
    int v,next;
}edge[MAXN];
void init()
{
    cot=0;
    memset(head,-1,sizeof(head));
    memset(vis,0,sizeof(vis));
    memset(in,0,sizeof(in));
}
void add(int u,int v)
{
    edge[++cot].next=head[u];
    edge[cot].v=v;
    head[u]=cot;
}
void dfs(int s)
{
    vis[s]=1;
    for(int i=head[s];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!vis[v])
            dfs(v);
    }
}

int a[MAXN];
int main()
{
    int ans1=0,ans2=0;
    cin>>n;
    init();
    for(int i=1;i<=n;i++)
        cin>>a[i];
    sort(a+1,a+n+1);
    a[0]=a[n+1]=-INF;
    if(n==1)
    {
        cout<<1<<endl;
       return 0;
    }
    for(int i=1;i<=n;i++)
    {
        if(abs(a[i]-a[i-1]) <= abs(a[i]-a[i+1]))
            add(i,i-1),in[i-1]++;
        else
            add(i,i+1),in[i+1]++;
    }
    for(int i=1;i<=n;i++)
    {
        if(!in[i])
        {
            dfs(i);
            ans1++;
        }
    }
    for(int i=1;i<=n;i++)
        if(!vis[i])
            ans2++;
    cout<<ans1+(ans2/2)<<endl;
    return 0;
}