CF 990B. Micro-World【数组操作/贪心/STL/二分搜索】

最新推荐文章于 2024-11-01 13:26:07 发布

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最新推荐文章于 2024-11-01 13:26:07 发布

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原文链接：http://www.cnblogs.com/Roni-i/p/9215028.html

本文探讨了一种算法问题，即如何通过特定条件下的元素删除操作来最小化数组中剩余元素的数量。通过对输入数组进行排序及使用二分查找等技术手段，文章提供了一种高效的解决方案。

【链接】：CF
【题意】：对任意一个数a[i] ，可以对任意满足 i != j 且 a[i] > a[j] && a[i] <= a[j] +k 的 a[j] 可以被删掉，求使最终剩下的个数最少。
【分析】：扫一遍，二分搜索合法的。
【代码】：

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
using namespace std;

typedef long long ll;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const ll LNF = 1e18;
const int maxn = 1e6+ 100;
const int maxm = 100;
const double PI = acos(-1.0);
const double eps = 1e-8;
//const int dx[] = {-1,1,0,0,1,1,-1,-1};
//const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dx[] = {-1,0,1,0};
int dy[] = {0,1,0,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int dir[6][3]={ {0,0,1},{0,0,-1},{-1,0,0},{1,0,0},{0,1,0},{0,-1,0} };
int n,k;
int a[maxn], b[maxn];
int main()
{
    while(cin>>n>>k)
    {
        int cnt=0;
        for(int i=0;i<n;i++) cin>>a[i];/*
        for(int i=0;i<n;i++)
        {
            cout<<i<<"   ";
        }
        cout<<endl;*/
        sort(a,a+n,greater<int>());/*
        for(int i=0;i<n;i++)
        {
            cout<<a[i]<<' ';
        }
        cout<<endl;*/
        for(int i=0;i<n;i++)
        {
            int p = lower_bound(a,a+n,a[i]+k,greater<int>())-a;   //第一个 <= (a[i]+k) 的位置
            //cout<<"a[i]+k="<<a[i]+k<<" p="<<p<<" a[p]="<<a[p]<<" a[i]="<<a[i];
            if(a[p]>a[i])
            {
                cnt++;
                //cout<<" YES";
            }
            //cout<<endl;
        }
        cout<<n-cnt<<endl;
    }
}
/*
7 1
101 53 42 102 101 55 54
42 53 54 55 101 101 102
outputCopy
3
42 55 102
inputCopy
6 5
20 15 10 15 20 25
25
outputCopy
1
inputCopy
7 1000000
1 1 1 1 1 1 1
outputCopy
7


7 1
101 53 42 102 101 55 54
101 53 42 102 101 55 54
102 101 101 55 54 53 42
a[i]+k=103 p=0 a[p]=102 a[i]=102
a[i]+k=102 p=0 a[p]=102 a[i]=101 YES
a[i]+k=102 p=0 a[p]=102 a[i]=101 YES
a[i]+k=56 p=3 a[p]=55 a[i]=55
a[i]+k=55 p=3 a[p]=55 a[i]=54 YES
a[i]+k=54 p=4 a[p]=54 a[i]=53 YES
a[i]+k=43 p=6 a[p]=42 a[i]=42
*/

[模拟]

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
using namespace std;

typedef long long ll;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const ll LNF = 1e18;
const int maxn = 1e6+ 100;
const int maxm = 100;
const double PI = acos(-1.0);
const double eps = 1e-8;
//const int dx[] = {-1,1,0,0,1,1,-1,-1};
//const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dx[] = {-1,0,1,0};
int dy[] = {0,1,0,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int dir[6][3]={ {0,0,1},{0,0,-1},{-1,0,0},{1,0,0},{0,1,0},{0,-1,0} };
int n,k,j;
int a[maxn], b[maxn];
int main()
{
    while(cin>>n>>k)
    {
        int cnt, j=0;
        for(int i=0;i<n;i++) cin>>a[i];

        sort(a,a+n);
        cnt=n;
        for(int i=0;i<n;i++)
        {
           while(a[j]<a[i])
           {
               if(a[j]+k>=a[i]) cnt--;
               j++;
           }
        }
        cout<<cnt<<endl;
    }
}

转载于:https://www.cnblogs.com/Roni-i/p/9215028.html