HDU 5087 (linear DP + secondary large LIS)

Source: Internet
Author: User

Question Link: Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 5087

Theme: Calculate the length of the second-largest Lis. Note that the two Lis with the same length are larger than the subscript and the larger Lis.

Solutions:

Struct records the maximum long fir of the current vertex, the second long sec.

The transfer of F [I]. FIR is actually a bare Lis.

But when f [J]. Fir + 1> = f [I]. fir, it also needs to be transferred. At this time, although the two LIS are of the same length, they are of different sizes.

For f [I]. Sec transfer, the first initial value is 0. Under the conditions of a [I]> A [J:

① When F [J]. Fir + 1> = f [I]. Fir:

We certainly have the maximum length of F [J]. Fir + 1, and the remaining values of F [I]. fir and f [J]. Sec + 1 are the same.

② When F [J]. fir + 1 <F [I]. at this time, the maximum length of the FIR is determined, but there is an additional optional value f [J]. fir + 1, note + 1

The HDU data is omitted, and no more than 1 can be used.

 

After the transfer of F [I] is completed, update the global result fir and SEC. I thought that the final result sec must be in all f [I]. Sec.

In fact, the SEC can also be f [I]. fir, which is a global growth.

If f [I]. Fir> = fir (still equal, although the length is the same)

At this time, there are three alternative answers: fir, F [I]. Sec, sec. Obviously, fir> = sec, so remove the SEC.

Update the longest fir.

 

#include "cstdio"#include "cstring"#include "iostream"using namespace std;int a[1005];struct status{    int fir,sec;    status() {}    status(int fir,int sec):fir(fir),sec(sec) {}}f[1005];int main(){    //freopen("in.txt","r",stdin);    int T,n;    scanf("%d",&T);    while(T--)    {        int fir=0,sec=0;        memset(f,0,sizeof(f));        scanf("%d",&n);        for(int i=1;i<=n;i++) scanf("%d",&a[i]);        for(int i=1;i<=n;i++)        {            f[i]=status(1,0);            for(int j=1;j<i;j++)            {                if(a[i]>a[j])                {                    if(f[j].fir+1>=f[i].fir)                    {                        f[i].sec=max(f[i].fir,f[j].sec+1);                        f[i].fir=f[j].fir+1;                    }                    else f[i].sec=max(f[i].sec,f[j].fir+1);                }            }            if(f[i].fir>=fir)            {                sec=max(f[i].sec,fir);                fir=f[i].fir;            }        }        printf("%d\n",sec);    }}

 

12046191 00:56:35 Accepted 5087 125 Ms 240 K 1099 B C ++ Physcal

HDU 5087 (linear DP + secondary large LIS)

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