hdu 5214 Movie(贪心)

Movie

Time Limit: 14000/7000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1390    Accepted Submission(s): 324

Problem Description

Cloud and Miceren like watching movies.

Today, they want to choose some wonderful scenes from a movie. A movie has N scenes can be chosen, and each scene is associate with an interval [L, R]. L is the beginning time of the scene and R is the ending time. However, they can't choose two scenes which have overlapping intervals. (For example, scene with [1, 2] and scene with [2, 3], scene with [2, 5] and scene with[3, 4]).

Now, can you tell them if they can choose such three scenes that any pair of them do not overlap?

Since there are so many scenes that you can't get them in time, we will give you seven parameters N, L1, R1, a, b, c, d, and you can generate L1 ~ LN, R1 ~ RN by these parameters.

 

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case contains seven integers N, L1, R1, a, b, c, d, meaning that there are N scenes. The i-th scene's interval is [Li, Ri]. L1 and R1 have been stated in input, and Li = (Li−1 ∗ a + b) mod 4294967296, Ri = (Ri−1 ∗ c + d) mod 4294967296.

After all the intervals are generated, swap the i-th interval's Li and Ri if Li > Ri.


T is about 100.

1 ≤ N ≤ 10000000.

1 ≤ L1,R1 ≤ 2000000000.

1 ≤ a,b,c,d ≤ 1000000000.

The ratio of test cases with N > 100 is less than 5%.

 

Output

For each test, print one line.

If they can choose such three scenes, output "YES", otherwise output "NO".

 

Sample Input

2
3 1 4 1 1 1 1
3 1 4 4 1 4 1

Sample Output

NO
YES

题目大意：给出一些区间，判断这些区间当中是否有至少三个不重叠的区间
题目分析：
很明显的贪心题，先找到最小的右边界，然后找到左边界比当前右边界大的区间中最小的右边界，如果能找够３次，那么说明是yes
比较坑的是，需要全部左右边界生成之后，再判断每一个边界是否要交换左右端点

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define MAX 10000007

using namespace std;


struct Node
{
    unsigned l,r;
}p[MAX];

unsigned n,a,b,c,d;
int t;

int main ( )
{
    scanf ( "%d" , &t );
    while ( t-- )
    {
        scanf ( "%u" , &n );
        scanf ( "%u%u" , &p[0].l , &p[0].r );
        scanf ( "%u%u%u%u" , &a , &b , &c , &d );
        for ( int i = 1 ; i < n ; i++ )
        {
            p[i].l = p[i-1].l*a + b;
            p[i].r = p[i-1].r*c + d;
            //if ( p[i].l > p[i].r ) swap ( p[i].l , p[i].r );
        }
        for ( int i = 0 ; i < n ; i++ )
            if ( p[i].l > p[i].r ) swap ( p[i].l , p[i].r );
        int flag = 0;
        unsigned cur = 0xffffffff, pos = -1;
        while ( flag < 3 )
        {
            cur = 0xffffffff;
            for ( int i = 0 ; i < n ; i++ )
                if ( p[i].l > pos || !flag )
                    cur = min ( cur , p[i].r );
           // cout << "CUr : " << cur << " " << pos << endl;
            if ( cur == 0xffffffff ) break;
            else pos = cur , flag++; 
           // cout << "YES : " << flag << endl;   
        }
        if ( flag == 3 ) puts ( "YES" );
        else puts ("NO");
    }
}