POJ-2533 longest Ordered subsequence (longest ascending subsequence)

D. Longest ascending sub-sequence

S. Attention is strictly incremented

C.O (NLOGN)

#include <iostream>#include<stdio.h>using namespacestd;Const intmaxn=1005;intA[MAXN],B[MAXN];//B[k] is the smallest end element value in sequence A of all increments of K//find a location using a binary search method to make num>b[i-1] and num<b[i] and use num instead of b[i]intSearch (intNumintLowintHigh ) {    intmid;  while(low<=High ) {Mid= (Low+high)/2; if(Num>=b[mid]) low=mid+1; Elsehigh=mid-1; }    returnLow ;}intDP (intN) {    intI,len,pos; b[1]=a[1]; Len=1;  for(i=2; i<=n;i++){        if(A[i]>b[len]) {//if A[i] is larger than the maximum size of the b[] array, insert it directly into the backlen=len+1; B[len]=A[i]; }        Else{//in the b[] array, find the first position larger than a[i] and let A[i] replace the position .Pos=search (A[i],1, Len); B[pos]=A[i]; }    }    returnLen;}intMain () {intN; inti;  while(~SCANF ("%d",&N)) {         for(i=1; i<=n;++i) {scanf ("%d",&A[i]); } printf ("%d\n", DP (N)); }    return 0;}

View Code

C2. O (n^2)

#include <iostream>#include<stdio.h>using namespacestd;#defineMAXN 1005intA[MAXN];intDP[MAXN];intMain () {intN; inti,j; inttemp; intans;  while(~SCANF ("%d",&N)) {         for(i=1; i<=n;++i) {scanf ("%d",&A[i]); } dp[1]=1;  for(i=2; i<=n;++i) {Temp=0;  for(j=1; j<i;++j) {                if(a[i]>A[j]) {                    if(temp<Dp[j]) {Temp=Dp[j]; }}} Dp[i]=temp+1; } ans=0;  for(i=1; i<=n;++i) {            if(dp[i]>ans) {ans=Dp[i]; }} printf ("%d\n", ans); }    return 0;}

View Code

POJ-2533 longest Ordered subsequence (longest ascending subsequence)