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====== 2020/05/24 牛客假日团队赛40 ====== ===== 比赛信息 ===== 日期： 2020/05/24 链接： [[https://ac.nowcoder.com/acm/contest/5207#description]]\\ 做题统计：王瑞琦：冯宇扬：1 常程：1\\ ===== 题解 ===== ==== A ==== solved by, upsolved by .\\ 题意：\\ 题解： ==== B ==== solved by, upsolved by .\\ 题意：\\ 题解：\\ ==== C ==== solved by, upsolved by .\\ 题意：\\ 题解： ==== D ==== solved by, upsolved by .\\ 题意：\\ 题解： ==== E Goldilocks and the N Cows ==== solved by 常程, upsolved by .\\ 题意：有N（$1\le N\le20000$）只奶牛，每只奶牛有一个舒适温度区间A(i)-B(i)，温度过低、温度过高、温度位于舒适区间都有不同产奶量，分别以XYZ表示，其中Y一定是最大的。每只牛的区间不同，求在某个温度下产奶量最大的值。\\ 题解：离散化区间边界。设定增量数组，在每只牛的边界处分别给增量数组加Y-X、Z-Y。初始奶量（温度很低）为3X，温度从小到大根据增量数组变化，得到最大值。


#include
#include
#include
using namespace std;
int n,x,y,z,a[20005],b[20005],line[40005],add[40005];
long long ans,now;
mapp;
int main()
{
    scanf("%d%d%d%d",&n,&x,&y,&z);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&a[i],&b[i]);
        line[2*i-1]=a[i];
        line[2*i]=++b[i];
    }
    sort(line+1,line+2*n+1);
    line[2*n+1]=-1;
    for(int i=1;i<=2*n;i++)
        if(line[i]!=line[i+1])
        {
            line[++line[0]]=line[i];
            p[line[i]]=line[0];
        }
    for(int i=1;i<=n;i++)
    {
        add[p[a[i]]]+=y-x;
        add[p[b[i]]]+=z-y;
    }
    ans=now=x*n;
    for(int i=1;i<=line[0];i++)
    {
        now+=add[i];
        ans=max(ans,now);
    }
    printf("%lld",ans);
    return 0;
}

==== F ==== solved by, upsolved by .\\ 题意：\\ 题解： ==== G ==== solved by, upsolved by .\\ 题意：\\ 题解： ==== H ==== solved by, upsolved by .\\ 题意：\\ 题解： ==== I ==== solved by 常程, upsolved by .\\ 题意：\\ 题解： ==== J ==== solved by, upsolved by .\\ 题意：\\ 题解： ==== K ==== solved by, upsolved by .\\ 题意：\\ 题解： ==== L ==== solved by, upsolved by .\\ 题意：\\ 题解： ===== replay ===== ===== 总结 =====