Beautiful People SGU, beautifulsgu

Beautiful People SGU, beautifulsgu

Longest ascending subsequence O (n log n): http://www.cnblogs.com/hehe54321/p/cf-340d.html

 

Title: https://cn.vjudge.net/problem/ZOJ-2319

Https://cn.vjudge.net/problem/SGU-199

A data type has two attributes: s and B. Now there are two instances x and y. Define if x. s <y. s & x. B <y. B or x. s> y. s & x. b> y. b, then x and y do not conflict, otherwise x and y conflict. Select the most data among the n data given, so that any two data does not conflict with each other.

Ideas:

The intuitive idea is to sort the original data by the first and second Keywords of s and B respectively, and then calculate the longest ascending subsequence by O (n log n. However, if the number of s in the first data is greater than that in the second data, the number of B in the first data is smaller than that in the second data, it is not certain whether the first or second data is good. Or, if a is not greater than B, a is not necessarily less than or equal to B. (It cannot be done anyway ...)

The correct method is to change it slightly. First, sort by s as the keyword, and then calculate the longest ascending subsequence according to B as the keyword. Of course, the longest ascending sub-sequence here requires s to be strictly less than, not just B Strictly less than, so more details need to be processed. The method used here is similar to this, is some tips http://blog.csdn.net/scnu_jiechao/article/details/40670393

 1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 struct P 5 { 6     int a1,a2,num; 7     bool operator<(const P& b)    const 8     { 9         return a1<b.a1||(a1==b.a1&&a2>b.a2);10     }11 };12 bool cmp(const P& a,const P& b)13 {14     return a.a2<b.a2;15 }16 P a[100100],s[100100];17 int f[100100],len,n,t;18 int main()19 {20     int i,j;21     scanf("%d",&n);22     for(i=1;i<=n;i++)23         scanf("%d%d",&a[i].a1,&a[i].a2),a[i].num=i;24     sort(a+1,a+n+1);25     for(i=1;i<=n;i++)26     {27         t=lower_bound(s+1,s+len+1,a[i],cmp)-s;28         s[t]=a[i];29         f[i]=t;30         len=max(len,t);31     }32     printf("%d\n",len);33     for(i=n,j=len;i>=1;i--)34         if(f[i]==j)35         {36             printf("%d ",a[i].num);37             j--;38         }39     return 0;40 }