Bridging signals Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6876 Accepted: 3728
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 third 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 functional 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 should 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 should 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
4 6 4 2 6 3 1 5 10 2 3 4 5 6 7 8 9 10 1 8 8 7 6 5 4 3 2 1 9 5 8 9 2 3 1 7 4 6Sample Output
3 9 1 4
刚开始没注意数据的规模，扫一眼题目就开始写，用的最经典的最长上升子序列的算法，结果返回超时，暴汗之下发现竟然有o(nlogn)算法， 主要思想就是用一个辅组数组b[]，用个len记录b的长度，（数组a[]记录输入数据），b[0]=a[0],len=1,用i依次循环a[]的元素，如果a[i]>b[len-1] 则b[len]=a[i],len++;否则找到b[]中第一个大于a[i]的元素b[j],b[j]=a[i];最后输出len，因为可以用二分搜索，所以算法为o(nlogn）的。 #include<iostream> #include<algorithm> #include<string> using namespace std; int main() { int i,j,t,n,len,start,end,mid; int a[40010],b[40010]; cin>>t; while(t--) { scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d",&a[i]); } len=1; b[0]=a[0]; for(i=1;i<n;i++) { if(a[i]>b[len-1]){b[len]=a[i];len++;} else if(len==1){b[len-1]=a[i];} else { start=0; end=len-1; while(start<=end) { mid=(start+end)/2; if(b[mid-1]<a[i]&&b[mid]>=a[i]){b[mid]=a[i];break;} else if(mid==0&&b[mid]>=a[i]){b[mid]=a[i];break;} else if(b[mid]>a[i])end=mid-1; else start=mid+1; } } } //for(i=0;i<len;i++){cout<<b[i]<<" ";} //cout<<endl; printf("%d/n",len); } return 0; }
