BZOJ2276: [Poi2011]Temperature

2276: [Poi2011]Temperature

Time Limit: 20 Sec  Memory Limit: 32 MB
Submit: 293  Solved: 117
[Submit][Status]

Description

The
Byteotian Institute of Meteorology (BIM) measures the air temperature
daily. The measurement is done automatically, and its result immediately
printed. Unfortunately, the ink in the printer has long dried out...
The employees of BIM however realised the fact only recently, when the
Byteotian Organisation for Meteorology (BOM) requested access to that
data.

An
eager intern by the name of Byteasar saved the day, as he
systematically noted down the temperatures reported by two domestic
alcohol thermometers placed on the north and south outside wall of the
BIM building. It was established decades ago by various BIM employees
that the temperature reported by the thermometer on the south wall of
the building is never lower than the actual temperature, while that
reported by the thermometer on the north wall of the building is never
higher than the actual temperature. Thus even though the exact
temperatures for each day remain somewhat of a mystery, the range they
were in is known at least.

Fortunately
for everyone involved (except Byteasar and you, perhaps), BOM does not
require exact temperatures. They only want to know the longest period in
which the temperature was not dropping (i.e. on each successive day it
was no smaller than on the day before). In fact, the veteran head of BIM
knows very well that BOM would like this period as long as possible. To
whitewash the negligence he insists that Byteasar determines, based on
his valuable notes, the longest period in which the temperature could have been
not dropping. Now this is a task that Byteasar did not quite expect on
his BIM internship, and he honestly has no idea how to tackle it. He
asks you for help in writing a program that determines the longest such
period.

某国进行了连续n天的温度测量，测量存在误差，测量结果是第i天温度在[l_i,r_i]范围内。
求最长的连续的一段，满足该段内可能温度不降。

Input

In
the first line of the standard input there is one integer
n(1<=N<=1000000) that denotes the number of days for which
Byteasar took notes on the temperature. The measurements from day are
given in the line no.i+1 Each of those lines holds two integers, x and y
(-10^9<=x<=y<=10^9). These denote, respectively, the minimum
and maximum possible temperature on that particular day, as reported by
the two thermometers.

In
some of the tests, worth 50 points in total, the temperatures never
drop below -50 degrees (Celsius, in case you wonder!) and never exceeds
50 degrees (-50<=x<=y<=50)

第一行n
下面n行，每行l_i，r_i
1<=n<=1000000

Output

In
the first and only line of the standard output your program should
print a single integer, namely the maximum number of days for which the
temperature in Byteotia could have been not dropping.

一行，表示该段的长度

Sample Input

6

6 10

1 5

4 8

2 5

6 8

3 5

Sample Output

4

HINT

 

Source

题解：

类似与pilots，我们可以枚举右端点 i，那么左端点 l[i]一定是单调不减的，那么就可以使用单调队列。

那么如何判断当前连续一段是否能单调不减呢？注意到如果x能被到达，那么所有y>x也一定能到达，而x就是这一段中温度最小值的最大值！

因为在到达该点之前，必须上升到x，之后又无法下降，所以合法的下界一定是这段区域里的温度最小的最大值，当然如果该值>当前i的上界，将队首元素弹出。

也就是说维护一个最小值单调递减的单调队列。

代码：

 #include<cstdio>

 #include<cstdlib>

 #include<cmath>

 #include<cstring>

 #include<algorithm>

 #include<iostream>

 #include<vector>

 #include<map>

 #include<set>

 #include<queue>

 #include<string>

 #define inf 1000000000

 #define maxn 1000000+5

 #define maxm 500+100

 #define eps 1e-10

 #define ll long long

 #define pa pair<int,int>

 #define for0(i,n) for(int i=0;i<=(n);i++)

 #define for1(i,n) for(int i=1;i<=(n);i++)

 #define for2(i,x,y) for(int i=(x);i<=(y);i++)

 #define for3(i,x,y) for(int i=(x);i>=(y);i--)

 #define mod 1000000007

 using namespace std;

 inline int read()

 {
     int x=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}

     return x*f;

 }
 int n,ans=,l=,r=,now,last=,a[maxn],b[maxn],q[maxn];

 int main()

 {

     freopen("input.txt","r",stdin);

     freopen("output.txt","w",stdout);

     n=read();
     for1(i,n)
     {
         a[i]=read();b[i]=read();
         while(l<=r&&a[q[r]]<=a[i])r--;
         q[++r]=i;
         now=last;
         while(a[q[l]]>b[i])now=q[l++]+;
         ans=max(ans,i-now+);
         last=now;
     }
     printf("%d\n",ans);

     return ;

 }