P3353 在你窗外闪耀的星星

Posted 2021-03-08 for-miku

链接：Miku

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去除题面，这就是一道线段树板子题，暴力枚举每一个区间然后去最大值就可以了

有星星重叠？本来就是在那个点加上去，继续加就可以了

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这就是道裸的区间和+单点修改

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以及，为了省事，我的线段树直接把右端点搞成了100000，毕竟一堆零没有任何影响

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 #include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m;
long long  ans;
long long sum[400005], lazy[400005];
int f,x,y;
long long k;
int rrr;
const int maxn=100000;
void pushdown(int x, int L, int R){
    if (lazy[x] != 0){
        int mid = (L + R) >> 1;
        lazy[x << 1] += lazy[x];
        lazy[x << 1 | 1] += lazy[x];
        sum[x << 1] += lazy[x] * (mid - L + 1);
        sum[x << 1 | 1] += lazy[x] * (R - mid);
        lazy[x] = 0;
    }
    return;
}
void pushup(int x){
    sum[x] = sum[x << 1] + sum[x << 1 | 1];
    return;
}
void update(int x, int l, int r, int L, int R, long long d){
    if (L <= l && r <= R){
        lazy[x] += d;
        sum[x] += d * (r - l + 1);
        return;
    }
    int mid = (l + r) >> 1;
    pushdown(x, l, r);
    if (L <= mid) update(x << 1, l, mid, L, R, d);
    if (R > mid) update(x << 1 | 1, mid + 1, r, L, R, d);
    pushup(x);
}
long long query(int x, int l, int r, int L, int R){
    if (L <= l && r <= R){
        return sum[x];
    }
    int mid = (l + r) >> 1;
    pushdown(x, l, r);
    long long ans = 0;
    if (L <= mid) ans += query(x << 1, l, mid, L, R);
    if (R > mid) ans += query(x << 1 | 1, mid + 1, r, L, R);
    return ans;
}
int main(){
    scanf("%d%d",&n,&m);
    if(m==0){
        cout<<0;
        return 0;
    }
    for(int i=1;i<=n;++i){
        scanf("%d%d",&x,&y);
        update(1, 1,maxn, x, x, y);
        rrr=max(rrr,x);
    }
    rrr-=m;
    rrr++;
    for(int i=1;i<=rrr;++i){
        ans=max(ans,query(1,1,maxn,i,i+m-1));
    }
    cout<<ans;
    return 0;
}

Ac