POJ 1631 Bridging signals (LIS binary fast method), pojbridging

POJ 1631 Bridging signals (LIS binary fast method), pojbridging

Language:DefaultBridging signals Time Limit:1000 MS   Memory Limit:10000 K Total Submissions:10762   Accepted:5899 Description 'Oh no, they 've done it again ', cries the chief designer at the Waferland chip factory. once more the routing designers have screwed up completely, making the signals on the chip connecting the ports of two functional blocks cross each other all over the place. at this late stage of the process, it is too expensive to redo the routing. instead, the engineers have to bridge the signals, using the th Ird dimension, so that no two signals cross. however, bridging is a complicated operation, and thus it is desirable to bridge as few signals as possible. the call for a computer program that finds the maximum number of signals which may be connected on the silicon surface without crossing each other, is imminent. bearing in mind that there may be thousands of signal ports at the boundary of a func Tional block, the problem asks quite a lot of the programmer. Are you up to the task? A typical situation is schematically depicted in figure 1. the ports of the two functional blocks are numbered from 1 to p, from top to bottom. the signal mapping is described by a permutation of the numbers 1 to p in the form of a list of p unique numbers in the range 1 to p, in which the I: th number specifies which port on the right side shocould be connected to the I: th port on the left side. two signals cross if and only if the straight lines connecting the two ports of each pair do. Input On the first line of the input, there is a single positive integer n, telling the number of test scenarios to follow. each test scenario begins with a line containing a single positive integer p <40000, the number of ports on the two functional blocks. then follow p lines, describing the signal mapping: On the I: th line is the port number of the block on the right side which shocould be connected to the I: th port of the block on the left side. Output For each test scenario, output one line containing the maximum number of signals which may be routed on the silicon surface without crossing each other. Sample Input 4642631510234567891018876543219589231746 Sample Output 3914 Source Northwestern Europe 2003

Because the data is too big, you must use a quick method. I 've been reading this for a long time today. Remember that c [I] represents the string with the length of I and the minimum value of the last bit.

Seach is to find the sequence ending with a [I], and then update

Code:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8using namespace std;#define N 100005int a[N],c[N],n;int seach(int len,int x){int le=0,ri=len,ans=0,mid;while(le<=ri){mid=(le+ri)>>1;if(c[mid]<x){ans=mid;le=mid+1;}        elseri=mid-1;}    return ans;}int main(){int i,j,t;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=1;i<=n;i++)scanf("%d",&a[i]);int len=1;c[1]=a[1];for(i=2;i<=n;i++){if(a[i]>c[len]){j=len++;}else  j=seach(len,a[i]);c[j+1]=a[i];}     printf("%d\n",len);}   return 0;}