Bridging signals
http://poj.org/problem?id=1631
http://acm.hdu.edu.cn/showproblem.php?pid=1950
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1986
Time Limit: 1000MS
Memory Limit: 10000K
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
46
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
39 1 4
介绍：
LIS(最长递增子序列)：有一列互不相等数arr[n]，从中取出k个数使得lis[k]单调递增，求k的最大值。
思路：
读一个数就计算一次。当它是第一个数时，我们就把它存到lis[1]；
当它不是第一个数时，就在lis中查找小于它且离它最近的数lis[i]，加到lis中(替换掉lis[i+1])，
找不到的话就说明它最小，我们替换掉lis[1]；
这样lis[n]的成为LIS的“潜力”就增加了。
注意两点：
1. 请使用二分查找，以防TLE。
2. 使用一个变量（比如len或count），作为lis中添加元素的个数（lis[0]不算），最后输出它就行。
方法代码：
/*POJ: 110ms,324KB*/
/*HDU: 141ms,548KB*/
/*ZOJ: 60ms,336KB*/
#include <cstdio>
#define MAXN 40002
using namespace std;
int arr[MAXN];
int find(int n, int num)
{
int l = 0, r = n, m;//left,right,middle
//二分查找
while (l <= r)
{
m = (l + r) >> 1;
arr[m] < num ? l = m + 1 : r = m - 1;
}
return l;
}
int main()
{
int n, l;
scanf("%d", &n);//n组数据
while (n--)
{
scanf("%d", &l);
int count = 1, id, num;
for (int i = 1; i <= l; i++)
arr[i] = MAXN;//防止在二分查找时出现越界查找返回异常结果
for (int i = 1; i <= l; i++)
{
scanf("%d", &num);
id = find(count, num);
if (count < id)
count = id;
arr[id] = num;
}
printf("%d\n", count);
}
return 0;
}
