SmallSXJ的博客

点击这里查看原题最终得到的一定是一个下凸壳，因此把所有直线以k升序为第一关键字，b降序为第二关键字进行排序。若当前直线能完全覆盖栈顶直线，即i与s[top]的交点在s[top]与s[top-1]的交点左侧，则将栈顶元素弹出。/*User:SmallLanguage:C++Problem No.:1007*/#include<bits/stdc++.h>#define ll long lo

点击这里查看原题单调栈模板题，注意答案要乘3。/*User:SmallLanguage:C++Problem No.:3039*/#include<bits/stdc++.h>#define ll long long#define inf 999999999using namespace std;const int M=1e3+5;int a[M][M],s[M],l[M],tp

点击这里查看原题维护一个单调递减的栈，如果要加入的元素x大于栈顶元素s[top]，分两种情况： 如果s[top-1]>x，ans+=x； 如果s[top-1]<=x，ans+=s[top-1]。/*User:SmallLanguage:C++Problem No.:1345*/#include<bits/stdc++.h>#define ll long long#define in

点击这里查看原题/*User:SmallLanguage:C++Problem No.:1113*/#include<bits/stdc++.h>#define ll long long#define inf 999999999using namespace std;const int M=25e4+5;int t,x,ans,n,tp,s[M];int main(){

点击这里查看原题水题/*User:SmallLanguage:C++Problem No.:1660*/#include<bits/stdc++.h>#define ll long long#define inf 999999999using namespace std;const int M=8e4+5;int n,c[M],s[M],tp;ll ans;int main(

点击这里查看原题做两次单调栈，栈里存序号/*User:SmallLanguage:C++Problem No.:1657*/#include<bits/stdc++.h>#define ll long long#define inf 999999999using namespace std;const int M=5e4+5;int n,h[M],c[M],ans,sum[M],

点击这里查看原题很简单的线段树，当然也有很多别的做法/*User:SmallLanguage:C++Problem No.:1012*/#include<bits/stdc++.h>#define ll long long#define inf 999999999#define lson (rt<<1)#define rson (lson|1)#define mid (l+r>>