题目描述
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。
输入输出格式
输入格式：
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)。
输出格式：
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。
输入输出样例
说明
【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。
题解
- 考虑一下二分出一个花盆的宽度，然后用单调对列判断是否满足时间差大于等于D
代码
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #define N 1000010 5 using namespace std; 6 int n,d,mx,maxv[N],minv[N],a[N],b[N]; 7 bool check(int x) 8 { 9 int headx=0,heady=0,tailx=0,taily=0; 10 for (int i=0;i<=mx;i++) 11 if (maxv[i]+1) 12 { 13 while (headx<tailx&&i-a[headx]>x) headx++; 14 while (heady<taily&&i-b[heady]>x) heady++; 15 while (headx<tailx&&maxv[i]>=maxv[a[tailx-1]]) tailx--; 16 while (heady<taily&&minv[i]<=minv[b[taily-1]]) taily--; 17 a[tailx++]=i,b[taily++]=i; 18 if (maxv[a[headx]]-minv[b[heady]]>=d) return true; 19 } 20 return false; 21 } 22 int main() 23 { 24 memset(maxv,-1,sizeof(maxv)),memset(minv,-1,sizeof(minv)); 25 scanf("%d%d",&n,&d); 26 for (int i=1,x,y;i<=n;i++) 27 { 28 scanf("%d%d",&x,&y),mx=max(mx,y); 29 if (maxv[x]==-1) maxv[x]=minv[x]=y; 30 maxv[x]=max(maxv[x],y),minv[x]=min(minv[x],y); 31 } 32 int l=1,r=mx+1; 33 while (l<r) 34 { 35 int mid=(l+r)>>1; 36 if (check(mid)) r=mid; else l=mid+1; 37 } 38 printf("%d",r==mx+1?-1:r); 39 }