Alternative Thinking（區間交集，貪心）

Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin’s string represents Kevin’s score on one of the n questions of the olympiad—‘1’ for a correctly identified cow and ‘0’ otherwise.

However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.

Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0’s in that substring to '1’s and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).

The following line contains a binary string of length n representing Kevin’s results on the USAICO.

Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.

Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring ‘10000011’ and turn his string into ‘10011011’, which has an alternating subsequence of length 5: ‘10011011’.

In the second sample, Kevin can flip the entire string and still have the same score.
題目分析：
通過這個題可以知道其實好多問題轉化一下都能轉化成好多基礎的題，無非就是細節多了點不同。
這個題沒必要吧問題和時間都恰的那麼準，如果第一個人溫度的變化區間和第二個人的溫度變化區間相交或包含，那麼第二個人必定滿意，如此類推，每更換一個人就更換一下區間的左右端點，左端點取最大值，右端點取最小值，因爲這樣是最準確的，如果不加上更換端點的操作的話左右端點就會無限的增加和減少，具體細節看代碼。

#include<iostream>
using namespace std;
int main()
{
    int q;
    cin>>q;
    while(q--)
    {
        int n,m;
        cin>>n>>m;
        int t,x,y;
        int l=m,r=m,time=0,flag=1;
        for(int i=1;i<=n;i++)
        {
            cin>>t>>x>>y;
            l-=(t-time);
            r+=(t-time);
            time=t;
            if(x>r||y<l) {flag=0;}//出現不滿意直接標記
            l=max(l,x);//更換左右端點
            r=min(r,y);
        }
        if(flag) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}