HDUOJ 1950 - Bridging signals（DP + 二分查找：最长递增子序列LIS【nlogn算法】）

原题：

Problem Description



……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.

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 6

Sample Output

3

9

1

4

解题思路：

nlogn优化算法解析可参考：

http://blog.csdn.net/shuangde800/article/details/7474903

https://www.cnblogs.com/sasuke-/p/5396843.html

代码：

最长递增子序列优化算法：O（nlogn）

#include <stdio.h>
int dp[40001], temp[40001];
int BinarySearch(int tail,int findValue)//二分查找
{
int head = 1;
int mid = (head + tail) / 2;
while (mid != 1 && mid != tail)
{
if (dp[mid] == findValue)return mid;
if (dp[mid] > findValue)
tail = mid;
else
head = mid;
mid = (head + tail) / 2;
if(head + 1 == tail && dp[mid] < findValue)
return mid + 1;
}
if (mid == 1 && dp[mid] > findValue)return 1;
else return 2;
}
int main() {
int T, N, i, dpIndex, j;
scanf("%d", &T);
while (T--) {
while (scanf("%d",&N)!=EOF)
{
scanf("%d", temp + 1);
dp[1] = temp[1];
dpIndex = 1;
for (i = 2; i <= N; i++)
{
scanf("%d", temp + i);
if(temp[i] > dp[dpIndex])
dp[++dpIndex] = temp[i];
else
dp[BinarySearch(dpIndex, temp[i])] = temp[i];
}
printf("%d\n", dpIndex);
}
}
}

最长递增子序列朴素算法：O（n^2）

#include<cstdio>
#include<cmath>
#include<cstring>
const int qq=1005;
int dp[qq];
int num[qq];
int main()
{
int n;
while(~scanf("%d",&n)){
memset(dp,0,sizeof(dp));
for(int i=0;i<n;++i)
scanf("%d",&num[i]);
dp[0]=1;
int x=0;
for(int j,i=0;i<n;++i){
int maxn=0;
for(j=0;j<i;++j)    //递推的原理
if(num[i]>num[j])    //由前面的最长递增子序列推出后面的最长递增子序列
maxn=maxn>dp[j]?maxn:dp[j];
dp[i]=maxn+1;
if(dp[i]>x)    x=dp[i];    //x记录的是当前的最大值、
}
printf("%d\n",x);
}
return 0;
}