Codevs 3641 God elect.

3641 God electtime limit: 1 sspace limit: 256000 KBtitle level: Golden GoldTitle Description Description

people all over the world have IQ IQ and the Emotional Quotient EQ . We use two numbers to denote the IQ and eq of a person, and the large number represents a higher IQ or EQ. Now there are n people in front of you , this n personal IQ and EQ are known, please choose as many as possible, asked to elect no two people i and J,i have an IQ greater than the IQ of J but I eq is less than J emotional quotient.

Enter a description Input Description

? first line a positive integer N , which represents the number of people. ? second line to article n+1 lines, two positive integers per line, respectively, representing each person's IQ and emotional quotient.

Output description Output Description

Only one row, for the maximum number of people selected.

Sample input Sample Input

3 the

Sample output Sample Output

<nobr>2 </nobr>

Data range and Tips Data Size & Hint

? n<=1000 ;  

/*basic idea: The IQ in descending order, and then according to this order, run the longest does not rise sub-sequence can be, nothing but the IQ of the high and low, as the original longest not rise sub-series array subscript just*/#include<iostream>using namespacestd; #include<cstdio>#defineN 1001intN; #include<algorithm>structpeo{intIq,love; intLine ;};intDp[n][n]; Peo Peo[n];voidinput () {scanf ("%d",&N);  for(intI=1; i<=n;++i) {scanf ("%d%d",&peo[i].iq,&peo[i].love); Peo[i].line=1; }}intCMP (Peo a,peo b) {returnA.iq>B.iq;}voidDP () {sort (peo+1, peo+n+1, CMP);  for(inti=n-1; i>=1;--i) for(intj=i+1; j<=n;++j)if(peo[i].love>=peo[j].love&&peo[j].line+1>peo[i].line) Peo[i].line=peo[j].line+1; intmaxx=0;  for(intI=1; i<=n;++i)if(peo[i].line>Maxx) Maxx=Peo[i].line; cout<<maxx<<Endl; }intMain () {input ();    DP (); return 0;}

View Code

Codevs 3641 God elect.