Figure 1. To the left: The two blocks' ports and their signal mapping (4,2,6,3,1,5). To the right: At most three signals may be routed on the silicon surface without crossing each other. The dashed signals must be bridged. 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 pecifies 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. InputOn 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. OutputFor 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 Input4642631510234567891018876543219589231746 Sample Output3914题目意思和输入的测试数据难理解,第一行输入一个数N代表几组测试数据,然后输入P代表每一组测试数据有几个数。比如第一组p=6输入的数据为4 2 6 3 1 5 依次与 1 2 3 4 5 6想对应,即连接起来,但是任何两条线都不能交叉,求最多的路的条数。#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int lis[40005];//存放longest increasing subsequence的数组int len;int Binary_Search(int n) //二分查找,时间复杂度为对数阶{ int low,high,mid; low = 1; high = len; while(low<high) { mid = (low+high)/2; if(lis[mid]>=n) high = mid; else low = mid+1; //只让下届做改变 } return low;}int main(){ int n,p,a; scanf("%d",&n); //n组测试数据 while(n--) { scanf("%d",&p); len = 0; lis[len]=0; while(p--) { scanf("%d",&a); if(a>lis[len]) { len++; lis[len]=a; } else { int pos = Binary_Search(a); lis[pos]=a; } } printf("%d/n",len); } return 0;}
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