Codeforces 528B Clique Problem dp+线段树(or 树状数组）

题目链接：点击打开链接

题意：

给定数轴上的n个点。

下面n行每行两个数 xi, wi 表示点和点权。

对于任意两个点u, v

若dis(u,v) >= u_w+v_w 则这两个点间可以建一条边。(in other words 若两点间距离大于两点的权值和则可以建边）

找一个最大团，输出这个最大团的点数。

其实对于一个权值点我们可以认为是一个区间

如: 4 5 ,可以认为是区间[-1, 9]

则这个点可以从区间(-inf, 1]转移过来，即val(9) = max(val(9), 区间(-inf, -1]最大值+1)

然后前面区间最值就用树状数组或者线段树维护

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <queue>
#include <set>
#include <vector>
#include <string>
#include <algorithm>
template <class T>
inline bool rd(T &ret) {
    char c; int sgn;
    if (c = getchar(), c == EOF) return 0;
    while (c != '-' && (c<'0' || c>'9')) c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return 1;
}
template <class T>
inline void pt(T x) {
    if (x <0) {
        putchar('-');
        x = -x;
    }
    if (x>9) pt(x / 10);
    putchar(x % 10 + '0');
}
using namespace std;
const int N = 400010;
#define L(x) tree[x].l
#define R(x) tree[x].r
#define Max(x) tree[x].max
#define Lson(x) (x<<1)
#define Rson(x) (x<<1|1)
#define Lazy(x) tree[x].lazy
struct Node{
    int l, r, max, lazy;
}tree[N<<2];
void Down(int id){
    Max(Lson(id)) = max(Lazy(id), Max(Lson(id)));
    Max(Rson(id)) = max(Lazy(id), Max(Rson(id)));
    Lazy(Lson(id)) = max(Lazy(id), Lazy(Lson(id)));
    Lazy(Rson(id)) = max(Lazy(id), Lazy(Rson(id)));
}
void U(int id){
    Max(id) = max(Max(Lson(id)), Max(Rson(id)));
}
void build(int l, int r, int id){
    L(id) = l; R(id) = r;
    Max(id) = Lazy(id) = 0;
    if (l == r)return;
    int mid = (l + r) >> 1;
    build(l, mid, Lson(id)); build(mid + 1, r, Rson(id));
}
int query(int l, int r, int id){
    if (l == L(id) && R(id) == r)return Max(id);
    Down(id);
    int mid = (L(id) + R(id)) >> 1, ans ;
    if (r <= mid)
        ans = query(l, r, Lson(id));
    else if (mid < l)
        ans = query(l, r, Rson(id));
    else ans = max(query(l, mid, Lson(id)), query(mid + 1, r, Rson(id)));
    U(id);
    return ans;
}
void up(int l, int r, int val, int id){
    if (l == L(id) && R(id) == r){
        Max(id) = max(Max(id), val);
        Lazy(id) = max(Lazy(id), val);
        return;
    }
    Down(id);
    int mid = (L(id) + R(id)) >> 1;
    if (r <= mid)
        up(l, r, val, Lson(id));
    else if (mid < l)
        up(l, r, val, Rson(id));
    else {
        up(l, mid, val, Lson(id));
        up(mid + 1, r, val, Rson(id));
    }
    U(id);
}
vector<int>G;
int hehehe;
struct Edge{
    int l, r;
}x[N];
bool cmp(Edge a, Edge b){
    return a.r < b.r;
}
int n, m;
int a[N], b[N];
int main(){
    while (~scanf("%d", &n)){
        G.clear();
        for (int i = 1; i <= n; i++){
            rd(a[i]); rd(b[i]);
            x[i].l = a[i] - b[i];
            x[i].r = a[i] + b[i];
            G.push_back(x[i].l);
            G.push_back(x[i].r);
        }
        sort(G.begin(), G.end());
        G.erase(unique(G.begin(), G.end()), G.end());
        for (int i = 1; i <= n; i++){
            x[i].l = lower_bound(G.begin(), G.end(), x[i].l) - G.begin() + 1;
            x[i].r = lower_bound(G.begin(), G.end(), x[i].r) - G.begin() + 1;
        }
        sort(x + 1, x + n + 1, cmp);
        build(1, G.size(), 1);
        int ans = 1;
        for (int i = 1; i <= n; i++){
            int tmp = query(1, x[i].l, 1) + 1;
            ans = max(ans, tmp);
            up(x[i].r, x[i].r, tmp, 1);
        }
        pt(ans); puts("");
    }
    return 0;
}