HDU 1950 Bridging signals（树状数组）

Bridging signals

Problem 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 rossing each other, is imminent. Bearing in mind that there may be housands 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?

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

4642631510234567891018876543219589231746

 

Sample Output

3914

【思路分析】
   还是和上一篇一样的题目，上一篇是用基于贪心和二分查找的算法求LIS，这篇是使用树状数组来求LIS。
   两个方法的思路有些类似，只不过树状数组是用来维护区间的和，个人觉得精髓还是差不多的。

代码如下：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn = 40005;int n,len;int a[maxn],c[maxn],val[maxn],pos[maxn];int lowbit(int x){    return x & (-x);}int sum(int i){    int ans = 0;    while(i > 0)    {        if(c[i] > ans)        {            ans = c[i];//直接赋值        }        i -= lowbit(i);    }    return ans;}void add(int i,int w){    while(i <= len)    {        if(w > c[i])        {            c[i] = w;//直接赋值        }        i += lowbit(i);    }}void init(){    scanf("%d",&n);    for(int i = 0;i < n;i++)    {        scanf("%d",&val[i]);        a[i] = val[i];    }    sort(a,a + n);//类似于二分法的数组d    len = unique(a,a + n) - a;//对数组a去重并得到去重后数组的长度    memset(c,0,sizeof(c));}void solve(){    int ans = 1,temp;    for(int i = 0;i < n;i++)    {        pos[i] = lower_bound(a,a + len,val[i]) - a + 1;//类似于二分法查找j的过程        temp = sum(pos[i] - 1) + 1;        if(temp > ans)            ans = temp;//更新最长子序列的长度        add(pos[i],temp);    }    printf("%d\n",ans);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        init();        solve();    }    return 0;}