CCPC2019江西省賽-Problem G.Traffic

題目描述：

　　/*純手打題面*/
　　Avin is observing the cars at a crossroads.He finds that there are n cars running in the east-west direction
with the i-th car passing the intersection at time a[i].There are another m cars running in the north-south
direction with the i-th car passing the intersection at time b[i].If two cars passing the intersections at 
the same time,a traffic crash occurs.In order to achieve world peace and harmony,all the cars running in the 
north-south direction wait the same time amount of integral time so that no two cars bump.You are asked the 
minimum waiting time.
/*純手打題面*/

Input

n m
a[i]
b[i]
(1<=n,m<=1000;1<=ai,bi<=1000)

Output

The minimum waiting time(integer).

題解心得：

啊啊啊啊啊啊啊啊啊啊啊啊，當時在比賽的時候翻了很久詞典，終於把這個題目的意思大概搞懂了

而後和咱們隊的大佬討論了很久

一開始爆搜O(N³)

而後想了很久奇奇怪怪的方法，分治什麼的，而後就剪枝優化。

可是最後用標記法把時間複雜度壓到了O(N²)。

大體思路：

枚舉時間（一個循環）

在每一個循環裏標記a[i]所佔的時間點（布爾數組c）

同時標記b[i]+t的時間點（布爾數組d）

接着c&&d;

就能夠了。

這題真的水，不加剪枝也能AC。

代碼：

之後附。

原文出處：https://www.cnblogs.com/XYYXP/p/JXCPC_2019_Problem_G.html