贪心

一个贪心算法总是做出当前最好的选择，也就是说，它期望通过局部最优选择从而得到全局最优的解决方案。—《算法导论》

区间问题

区间选点

20250805214625

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using namespace std;

const int N = 100010;

int n;
struct Range{
    int l,r;
    bool operator <(const Range& W)const{
        return r<W.r;
    }//重载小于号
}range[N];

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ){
        int l,r;
        scanf("%d%d", &l, &r);
        range[i]={l,r};//读入l,r
    }
    sort(range,range+n);//按右端点进行排序
    int res=0,ed=-2e9;//ed代表上一个点的右端点
    for (int i = 0; i < n; i ++ ){
        if(range[i].l>ed){
            res++;//点的数量加一
            ed=range[i].r;
        }
    }
    printf("%d\n",res);
    return 0;
}

最大不相交区间数量

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using namespace std;

const int N = 100010;

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

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ){
        int l,r;
        scanf("%d%d", &l, &r);
        range[i]={l,r};
    }
    sort(range,range+n);
    int res=0,ed=-2e9;
    for (int i = 0; i < n; i ++ ){
        if(range[i].l>ed){
            res++;
            ed=range[i].r;
        }
    }
    printf("%d\n",res);
    return 0;
}

区间分组

20250805214651

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using namespace std;

const int N = 1e5+10;

int n;
struct Range{
    int l,r;
    bool operator<(const Range &W)const{
        return l<W.l;
    }//按左端点排序
}range[N];

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ){
        int l,r;
        scanf("%d%d", &l, &r);
        range[i]={l,r};
    }
    sort(range,range+n);//sort排序
    priority_queue<int,vector<int>,greater<int>> heap;//小根堆维护所有组的右端点最小值
    for (int i = 0; i < n; i ++ ){//从左往右枚举
        auto r=range[i];//选择当前区间
        if(heap.empty()||heap.top()>=r.l)heap.push(r.r);
        else{
            heap.pop();
            heap.push(r.r);
        }
    }
    printf("%d\n",heap.size());
    return 0;
}