【扫描线】 codeforces 391D2 Supercollider

注意到没有两条共线的线段具有公共点，没有重合的线段说明每个十字形最多涉及一个水平线段和一个竖直线段这说明我们可以二分了:每条线段两端砍掉delta长度后，有没有线段有公共点？很水的扫描线吧~  按 X 轴扫描，先把横的线段扫进去，再用竖的线段去查询但是写法是很重要的，我说我不用线段树你信喵？我说我连离散化都不用你信喵？只存操作(包括插入、删除、询问)真是无比的高大上，跪跪跪还有那鬼畜的查询写法(set).lower_bound(l)!=(set).upper_bound(r)可以查询 [l..r] 中是否有数存在，至于为什么这样是对的而不是别的样子了，大家在纸上画画就知道了吧其实更好理解的是当 (set).lower_bound(l)==(set).upper_bound(r) 时， [l..r] 中一定没有数lower_bound是为了将 l 算入 [l..r] 中，upper_bound(r) 也是为了将 r 算入 [l..r] 中#include <iostream>  #include <queue>  #include <stack>  #include <map>  #include <set>  #include <bitset>  #include <cstdio>  #include <algorithm>  #include <cstring>  #include <climits>  #include <cstdlib>//#include <cmath>#include <time.h>#define maxn 300005#define maxm 100005#define eps 1e-10#define mod 1000000007#define INF 999999999#define lowbit(x) (x&(-x))#define mp mark_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid  #define rson o<<1 | 1, mid+1, R  typedef long long LL;//typedef int LL;using namespace std;LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}// headstruct point{int a, b, h;}line[maxn];struct node{int k, h, x;}op[maxn];set<int> s;int x1[maxn], x2[maxn], y1[maxn], y2[maxn];int n, m1, m2;void read(void){/*scanf("%d", &n);for(int i = 1; i <= n; i++) {scanf("%d%d%d%d", &x1[i], &y1[i], &x2[i], &y2[i]);if(x1[i] == x2[i]) {if(y1[i] > y2[i]) swap(y1[i], y2[i]);}else {if(x1[i] > x2[i]) swap(x1[i], x2[i]);}}*/scanf("%d%d", &m1, &m2), n = m1+m2;for(int i = 1; i <= m1; i++) {scanf("%d%d%d", &x1[i], &y1[i], &y2[i]), x2[i] = x1[i], y2[i] += y1[i];if(y1[i] > y2[i]) swap(y1[i], y2[i]);}for(int i = m1+1; i <= n; i++) {scanf("%d%d%d", &x1[i], &y1[i], &x2[i]), y2[i] = y1[i], x2[i] += x1[i];if(x1[i] > x2[i]) swap(x1[i], x2[i]);}}int cmp1(node a, node b){if(a.h == b.h) return a.k < b.k;else return a.h < b.h;}int cmp2(point a, point b){return a.h < b.h;}bool check(int x){m1 = m2 = 0;for(int i = 1; i <= n; i++) {if(x1[i] == x2[i]) {if(y2[i] - y1[i] >= 2*x) {op[m1].k = 1, op[m1].x = x1[i], op[m1++].h = y1[i]+x;op[m1].k = -1, op[m1].x = x1[i], op[m1++].h = y2[i]-x+1;}}else {if(x2[i] - x1[i] >= 2*x) {line[m2].a = x1[i]+x;line[m2].b = x2[i]-x;line[m2++].h = y1[i];}}}sort(op, op+m1, cmp1);sort(line, line+m2, cmp2);s.clear();for(int i = 0, j = 0; i < m2; i++) {while(j < m1 && line[i].h >= op[j].h)if(op[j].k == 1) s.insert(op[j++].x);else s.erase(op[j++].x);if(s.lower_bound(line[i].a) != s.upper_bound(line[i].b)) return true;}return false;}void work(void){int bot = 1, top = 1000000000, mid, res = -1;while(top >= bot) {mid = (top + bot) >> 1;if(check(mid)) bot = mid+1, res = mid;else top = mid-1;}if(~res) printf("%d\n", res);else printf("0\n");}int main(void){read();work();return 0;}