贪心算法-区间问题 C++

发布于:2024-11-27 ⋅ 阅读:(110) ⋅ 点赞:(0)

题目一

在这里插入图片描述

解题思路

原题解:https://www.acwing.com/solution/content/79913/在这里插入图片描述

代码实现

#include<iostream>
#include<algorithm>

using namespace std;

const int N = 1e5 + 10;

struct Range {
    int l, r;
    
    bool operator < (const Range &w) const {
        return r < w.r;
    }
}range[N];

int main()
{
    int n;
    cin >> n;
    
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d%d", &range[i].l, &range[i].r);
    }
    
    sort(range, range + n);
    
    int res = 0, ed = -0x3f3f3f3f;
    for (int i = 0; i < n; i ++ )
    {
        if (ed < range[i].l)
        {
            res ++;
            ed = range[i].r;
        }
    }
    
    cout << res;
    return 0;
}

题目二

在这里插入图片描述

解题思路

在这里插入图片描述

代码实现

#include<iostream>
#include<algorithm>

using namespace std;

const int N = 1e5 + 10;

struct Range {
    int l, r;
    bool operator < (const Range &w) const {
        return r < w.r;
    }
}range[N];

int main()
{
    int n;
    cin >> n;
    
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d%d", &range[i].l, &range[i].r);
    }
    
    sort(range, range + n);
    int res = 0, ed = -0x3f3f3f3f;
    
    for (int i = 0; i < n; i ++ )
    {
        if (ed < range[i].l)
        {
            ed = range[i].r;
            res ++;
        }
    }
    
    cout << res;
    return 0;
}

题目三

在这里插入图片描述

解题思路

原题解:https://www.acwing.com/solution/content/14773/
在这里插入图片描述

代码实现

#include<iostream>
#include<algorithm>
#include<queue>

using namespace std;

const int N = 1e5 + 10;

struct Range {
    int l, r;
    bool operator < (const Range &w) const
    {
        return l < w.l;
    }
}range[N];

int main()
{
    int n;
    cin >> n;
    
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d%d", &range[i].l, &range[i].r);
    }
    
    sort(range, range + n);
    priority_queue<int ,vector<int>, greater<int>> heap;
    
    for (int i = 0; i < n; i ++)
    {
        if (heap.empty() || range[i].l <= heap.top())
        {
            heap.push(range[i].r);
        }
        else
        {
            heap.pop();
            heap.push(range[i].r);
        }
    }
    
    cout << heap.size();
    return 0;
}

题目四

在这里插入图片描述

解题思路

原题解:https://www.acwing.com/solution/content/16980/
在这里插入图片描述
为什么r = -2e9不能放在for循环内:
例如样例:
4 10 2
4 5
11 12
当第一轮st更新后是5,第二轮j 还是 0,不过此时应该退出了,但如果-2e9放在外面,
if (r < st)
{
res = -1;
break;
}
就不会执行

代码实现

#include<iostream>
#include<algorithm>

using namespace std;

const int N = 1e5 + 10;

struct Range {
    int l, r;
    bool operator < (const Range &w) const {
        return l < w.l;
    }
}range[N];

int main()
{
    int st, ed, n;
    cin >> st >> ed >> n;
    
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d%d", &range[i].l, &range[i].r);
    }
    
    sort(range, range + n);
    
    int res = 0;
    bool flag = false;
    for (int i = 0; i < n; i ++)
    {
        int j = i, r = -0x3f3f3f3f;
        while (range[j].l <= st && j < n)
        {
            r = max(r, range[j].r);
            j ++;
        }
        
        if (r < st)
        {
            flag = true;
            break;
        }
        
        res ++;
        st = r;
        
        if (r >= ed)
        {
            break;
        }
        
        i = j - 1;
    }
    
    if (flag || st < ed)
    {
        res = -1;
    }
    
    cout << res;
    return 0;
}

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