HDU 5214 MOVIE

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5214

题面：

Movie

Time Limit: 14000/7000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 967    Accepted Submission(s): 239

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

23 1 4 1 1 1 13 1 4 4 1 4 1

 

Sample Output

NOYES

 

Source

赛码"BestCoder"杯中国大学生程序设计冠军赛

 

题意：是否能找到三个区间，互不交叉。

解题：开始没想到贪心，后来仔细一想还是挺清晰的，取两头最边上的，中间遍历，这样都找不到可行解的话，自然就无解了。排序实际上是没有必要的，排个序都O(nlogn)了，遍历一遍只需O（n）。

坑点：

1.l,r可能开始就大小相反

2.全部生成完数据后，才交换。

代码：

#include <iostream>#include <algorithm>using namespace std;struct info{    unsigned int l,r;}store[10000005];unsigned MAX(unsigned a,unsigned b){  if(a>b)return a;  else return b;}unsigned MIN(unsigned a,unsigned b){  if(a<b)return a;  else return b;}int main(){    unsigned t,n,le,ri,a,b,c,d,tmp,cnt,minri,maxle;    cin>>t;    bool flag;    while(t--)    {        cnt=0;        flag=false;        cin>>n>>le>>ri>>a>>b>>c>>d;        if(le>ri)        {            minri=le;            maxle=ri;        }        else        {          minri=ri;          maxle=le;         }        store[cnt].l=le;        store[cnt].r=ri;        cnt++;        for(unsigned i=1;i<n;i++)        {            le=(le*a + b);            ri=(ri*c + d);             tmp=MIN(le,ri);            if(tmp>maxle)            maxle=tmp;            tmp=MAX(le,ri);            if(tmp<minri)            minri=tmp;            store[cnt].l=le;            store[cnt].r=ri;            cnt++;        }        for(int i=0;i<cnt;i++)        {          le=MIN(store[i].l,store[i].r);          ri=MAX(store[i].l,store[i].r);            if(le>minri&&ri<maxle)            {                flag=true;                break;            }        }        if(flag)cout<<"YES\n";        else cout<<"NO\n";    }    return 0;}