自拉比赛B题（区间更新）

DescriptionKefa wants to celebrate his first big salary by going to restaurant. However, he needs company.Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factorin respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d unitsof money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!InputThe first line of the input contains two space-separated integers, n and d (1 ≤ n ≤ 105,)— the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line containsthe description of the i-th friend of type mi, si (0 ≤ mi, si ≤ 109)— the amount of money and the friendship factor, respectively.OutputPrint the maximum total friendship factir that can be reached.Sample InputInput

4 5
75 5
0 100
150 20
75 1

Output

100

Input

5 1000 711 3299 1046 887 54

Output

111

题解：区间更新思路。题比较水。

直接看代码吧：

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<iostream>#include<math.h>using namespace std;const int maxn=100010;struct py//定义一个结构体数组，赋予其中两个含义：金钱和人品{long long qian;long long renpin;} flog[maxn];bool cmp(py x,py y)//定义升序排列金钱函数{return x.qian<y.qian;}int main(){int n,m;long long s[maxn];//存储可加人品区间的总值long long maxx=0;//最大的人品while(~scanf("%d%d",&n,&m)){for(int i =1; i<=n; i++){scanf("%lld",&flog[i].qian);//赋值金钱scanf("%lld",&flog[i].renpin);//赋值人品}sort(flog+1,flog+n+1,cmp);//调用cmp进行金钱排序memset(s,0,sizeof(s));int k=1;for(int j=1; j<=n; j++){s[j]=s[j-1]+flog[j].renpin;//求可加区间人品总值maxx=max(maxx,flog[j].renpin);//求出最大的人品while(flog[j].qian-flog[k].qian>=m)//若不符合金钱差，更新可加区间第一个值{k++;}maxx=max(maxx,s[j]-s[k-1]);//求出最终的最大人品值}printf("%lld",maxx);//输出结果}return 0;}