标签：++ 大小 mes pen getc axis rds nbsp 结合
Farmer John has been having trouble making his plants grow, and needs your help to water them properly. You are given the locations of N raindrops (1 <= N <= 100,000) in the 2D plane, where y represents vertical height of the drop, and x represents its location over a 1D number line:
Each drop falls downward (towards the x axis) at a rate of 1 unit per second. You would like to place Farmer John‘s flowerpot of width W somewhere along the x axis so that the difference in time between the first raindrop to hit the flowerpot and the last raindrop to hit the flowerpot is at least some amount D (so that the flowers in the pot receive plenty of water). A drop of water that lands just on the edge of the flowerpot counts as hitting the flowerpot.
Given the value of D and the locations of the N raindrops, please compute the minimum possible value of W.
老板需要你帮忙浇花。给出N滴水的坐标，y表示水滴的高度，x表示它下落到x轴的位置。
每滴水以每秒1个单位长度的速度下落。你需要把花盆放在x轴上的某个位置，使得从被花盆接着的第1滴水开始，到被花盆接着的最后1滴水结束，之间的时间差至少为D。
我们认为，只要水滴落到x轴上，与花盆的边沿对齐，就认为被接住。给出N滴水的坐标和D的大小，请算出最小的花盆的宽度W。
第一行2个整数 N 和 D。
第2.. N+1行每行2个整数，表示水滴的坐标(x,y)。
仅一行1个整数，表示最小的花盆的宽度。如果无法构造出足够宽的花盆，使得在D单位的时间接住满足要求的水滴，则输出-1。
4 5 6 3 2 4 4 10 12 15
2
【样例解释】
有4滴水， (6,3), (2,4), (4,10), (12,15).水滴必须用至少5秒时间落入花盆。花盆的宽度为2是必须且足够的。把花盆放在x=4..6的位置，它可以接到1和3水滴, 之间的时间差为10-3 = 7满足条件。
【数据范围】
40%的数据：1 ≤ N ≤ 1000，1 ≤ D ≤ 2000；
100%的数据：1 ≤ N ≤ 100000，1 ≤ D ≤ 1000000，0≤x,y≤10^6。
看数据范围，结合题意，求高度差大于等于d的最短距离
单调队列
枚举结束的点
用两个队列分别维护最大值与最小值
求差更新答案便好
1 /************************** 2 User£ºMandy.H.Y 3 Language:c++ 4 Problem£º 5 **************************/ 6 7 #include<bits/stdc++.h> 8 9 using namespace std; 10 11 const int maxn = 1e5 + 5; 12 const int maxd = 1e6 + 5; 13 14 int n,d; 15 int l1,r1,q1[maxn]; 16 int l2,r2,q2[maxn]; 17 18 struct Node{ 19 int x,y; 20 }node[maxn]; 21 22 template<class T>inline void read(T &x){ 23 x = 0;bool flag = 0;char ch = getchar(); 24 while(!isdigit(ch)) flag |= ch == ‘-‘,ch = getchar(); 25 while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48),ch = getchar(); 26 if(flag) x = -x; 27 } 28 29 template<class T>void putch(const T x){ 30 if(x > 9) putch(x / 10); 31 putchar(x % 10 | 48); 32 } 33 34 template<class T>void put(const T x){ 35 if(x < 0) putchar(‘-‘),putch(-x); 36 else putch(x); 37 } 38 39 void file(){ 40 freopen("1232.in","r",stdin); 41 freopen("1232.out","w",stdout); 42 } 43 44 bool cmp(const Node &a,const Node &b){ 45 return a.x < b.x; 46 } 47 48 void readdata(){ 49 read(n);read(d); 50 for(int i = 1;i <= n; ++ i){ 51 read(node[i].x); 52 read(node[i].y); 53 } 54 sort(node + 1,node + n + 1,cmp); 55 } 56 57 void work(){ 58 int ans = node[n].x + 100; 59 for(int i = 1;i <= n; ++ i){ 60 while(l1 < r1 && node[i].x-node[q1[l1]].x>=ans) l1++; 61 while(l2 < r2 && node[i].x-node[q2[l2]].x>=ans) l2++; 62 if(l1 < r1 && abs(node[i].y-node[q1[l1]].y) >= d) ans = min(ans,node[i].x-node[q1[l1]].x); 63 if(l2 < r2 && abs(node[i].y-node[q2[l2]].y) >= d) ans = min(ans,node[i].x-node[q2[l2]].x); 64 while(l1 < r1 && node[i].y < node[q1[r1-1]].y) r1--; 65 while(l2 < r2 && node[i].y < node[q2[r2-1]].y) r2--; 66 q1[r1++] = i; 67 q2[r2++] = i; 68 } 69 if(ans > node[n].x - node[1].x) puts("-1"); 70 else put(ans); 71 } 72 73 int main(){ 74 // file(); 75 readdata(); 76 work(); 77 return 0; 78 }
luogu 2698 [USACO12MAR]花盆Flowerpot
标签：++ 大小 mes pen getc axis rds nbsp 结合
原文地址：https://www.cnblogs.com/Mandy-H-Y/p/11494536.html
