【模版】区间问题

区间问题

905. 区间选点

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5+10;
struct Node {
    int l, r;
    bool operator < (const Node &t ) const {
        return r < t.r;
    }
}segs[N];
int n;

int main() {
    cin >> n;
    
    int l, r;
    for(int i=0; i<n; i++) {
        cin >> l >> r;
        segs[i] = {l, r};
    }
    
    sort(segs, segs+n);
    
    
    int res=0, ed=-1e10;
    for(int i=0; i<n; i++) {
        if(segs[i].l > ed) {
            res++;
            ed = segs[i].r;
        }
    }
    
    cout << res;
    
    
    return 0;
}

---

908. 最大不相交区间数量

#include <bits/stdc++.h>
using namespace std;


const int N = 100010;
struct seg{
    int l, r;
    bool operator < (const seg &t) const {
        return r < t.r;
    }
}segs[N];

int main() {
    
    int n;
    cin >> n;
    for(int i=0; i<n; i++) {
        int l, r;
        cin >> l >> r;
        segs[i] = {l ,r};
    }
    
    sort(segs, segs+n);
    
    int res = 0, l = -2e9;
    
    for(int i=0; i<n; i++) {
        if(l < segs[i].l) {
            res++;
            l = segs[i].r;
        }
    }
    
    cout << res;
    
    return 0;
}

906. 区间分组

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5+10;
int n;
struct seg{
    int l, r;
    bool operator < (const seg &t) const {
        return l < t.l;
    }
}segs[N];

int main() {
    cin >> n;
    for(int i=0; i<n; i++) {
        int l, r;
        cin >> l >> r;
        segs[i] = {l, r};
    }
    
    sort(segs, segs+n);
    
    priority_queue<int, vector<int>, greater<int> > heap;
    for(int i=0; i<n; i++) {
        
        //如果组中最小的组的右边都大于新加入线段的左边，说明有交集，需要新开组
        if(heap.empty() || heap.top() >= segs[i].l) heap.push(segs[i].r);
        
        else {
            //否则加入到其中一个组中
            heap.pop();
            heap.push(segs[i].r);
        }
    }
    
    cout << heap.size();
    
    
    return 0;
}

907. 区间覆盖

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5+10;
struct seg {
    int l, r;
    bool operator < (const seg &t) const {
        return l < t.l;
    }
}segs[N];
int n;

int main() {
    int st, ed;
    
    cin >> st >> ed >> n;
    
    for(int i=0; i<n; i++) {
        int l, r;
        cin >> l >> r;
        segs[i] = {l, r};
    }
    
    sort(segs, segs+n);
    
    int res = 0;
    bool success = false;
    for(int i=0; i<n; i++) {
        int j = i, r = -2e9;
        while(j<n && segs[j].l <= st) {
            r = max(r, segs[j].r);
            j++;
        }
        
       
        if(r < st) {
            res = -1;
            break;
        }
        
        res++;
        if(r>=ed) {
            success = true;
            break;
        }
        
        st = r;
        i = j-1;
    }
    
    if(success) cout << res;
    else cout << -1;
    
    return 0;
}

Huffman树

148. 合并果子

#include <bits/stdc++.h>
using namespace std;

const int N = 10010;
int n;

int main() {
    cin >> n;
    priority_queue<int, vector<int>, greater<int> > heap;
    
    while(n--) {
        int x;
        cin >> x;
        heap.push(x);
    }
    
    int res = 0;
    while(heap.size()>1) {
        int a = heap.top(); heap.pop();
        int b = heap.top(); heap.pop();
        res += a + b;
        heap.push(a+b);
    }
    
    cout << res;
    
    return 0;
}

排序不等式

913. 排队打水

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 1e5+10;
int n;
int a[N];

int main() {
    
    cin >> n;
    for(int i=0; i<n; i++) scanf("%d", &a[i]);
    
    sort(a, a+n);
    
    LL res = 0;
    for(int i=0; i<n; i++) res += a[i]*(n-i-1);
    
    
    cout << res;
    
    
    return 0;
}

绝对值不等式

104. 货仓选址

推公式

125. 耍杂技的牛