HDU 4007 Dave （线段树扫描线 或 暴力+双扫描线）

【题意】给了n个点，求一个正方形能围住的最大点数，同样正方形平行于坐标轴。

【解题方法1】线段树+扫描线。AC时间：30ms。

////Created by just_sort 2016/12/3//Copyright (c) 2016 just_sort.All Rights Reserved//#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>#include <ext/pb_ds/hash_policy.hpp>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;using namespace __gnu_pbds;typedef long long LL;typedef pair<int, LL> pp;#define MP(x,y) make_pair(x,y)const int maxn = 10010;typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;//headLL L, R, n, r;LL b[3005];struct seg{    LL x, y, v;    seg(){}    seg(LL x, LL y, LL v):x(x), y(y), v(v){}}s[3005];struct node{    int l, r;    LL num, lazy;}T[30005];bool cmp1(const seg &a, const seg &b){    if(a.x == b.x) return a.v > b.v;    return a.x < b.x;}bool cmp2(const seg &a, const seg &b){    return a.y < b.y;}void pushup(int rt){    T[rt].num = max(T[rt*2].num, T[rt*2+1].num);}void pushdown(int rt){    if(T[rt].lazy != 0)    {        T[rt*2].lazy += T[rt].lazy;        T[rt*2+1].lazy += T[rt].lazy;        T[rt*2].num += T[rt].lazy;        T[rt*2+1].num += T[rt].lazy;        T[rt].lazy = 0;    }}void Build(int l, int r, int rt){    T[rt].l = l, T[rt].r = r, T[rt].num = T[rt].lazy = 0;    if(l == r) return ;    int mid = (l + r) / 2;    Build(l, mid, rt*2);    Build(mid+1, r, rt*2+1);}void update(int v, int rt){    if(L <= T[rt].l && T[rt].r <= R){        T[rt].lazy += v;        T[rt].num += v;        return ;    }    pushdown(rt);    int mid = (T[rt].l + T[rt].r) / 2;    if(L <= mid) update(v, rt*2);    if(mid < R) update(v, rt*2+1);    pushup(rt);}int main(){    while(scanf("%I64d%I64d",&n,&r)!=EOF)    {        memset(b, 0, sizeof(b));        for(int i = 1; i <= n; i++){            LL x, y;            scanf("%I64d%I64d",&x, &y);            s[3*i - 2] = seg(x, y, 1);            s[3*i - 1] = seg(x+r, y, -1);            s[3*i] = seg(x, y+r, 0);        }        sort(s+1, s+3*n+1, cmp2);        for(int i = 1; i <= 3*n; i++) b[i] = s[i].y;        int sz = unique(b+1, b+3*n+1) - b - 1;        sort(s+1, s+3*n+1, cmp1);        Build(1, sz, 1);        LL ans = 0;        for(int i = 1; i <= 3*n; i++){            if(s[i].v == 0) continue;            L = lower_bound(b + 1, b + sz + 1, s[i].y) - b;            R = lower_bound(b + 1, b + sz + 1, s[i].y + r) - b;            update(s[i].v, 1);            ans = max(ans, T[1].num);        }        printf("%lld\n", ans);    }    return 0;}

【解题方法2】 双扫描线+暴力  AC时间：780ms

【具体步骤】首先将所有点按横坐标大小排序，然后以a[i].x为第一条竖扫描线，a[i].x+r为第二条竖扫描线，在存下在这个区间内的点的y坐标b[i]。将b数组从小到大排序，再以b[i]为第一条横扫描线，b[i]+r为第二条横扫描线，如果寻找到满足的坐标，curans++。最后找出最大的nexans，记得下一次操作curans前重新赋值!

【AC代码】

////Created by just_sort 2016/12/3//Copyright (c) 2016 just_sort.All Rights Reserved//#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>#include <ext/pb_ds/hash_policy.hpp>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;using namespace __gnu_pbds;typedef long long LL;typedef pair<int, LL> pp;#define MP(x,y) make_pair(x,y)const int maxn = 10010;typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;//headstruct node{    int x, y;    node(){}    node(int x, int y) : x(x), y(y) {}    bool operator<(const node &rhs) const{        if(x == rhs.x) return y < rhs.y;        return x < rhs.x;    }}a[1010];int b[1010];int main(){    int n, r;    while(scanf("%d%d",&n,&r) != EOF)    {        for(int i = 1; i <= n; i++) scanf("%d%d",&a[i].x,&a[i].y);        sort(a+1, a+n+1);        int curans = 0, nexans = 0;        for(int i = 1; i <= n; i++){            int xUp = a[i].x + r;            int k = 0;            for(int j = i; j <= n; j++){                if(a[j].x <= xUp){                    b[++k] = a[j].y;                }            }            sort(b+1, b+k+1);            for(int j = 1; j <= k; j++){                int yUp = b[j] + r;                curans = 0;                for(int l = j; l <= k; l++){                    if(b[l] <= yUp){                        curans++;                    }                }                nexans = max(nexans, curans);            }        }        printf("%d\n",nexans);    }    return 0;}