HDU 1423 Greatest Common Increas…

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1065    Accepted Submission(s): 319

Problem Description

This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.

 

Input

Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.

 

Output

output print L - the length of the greatest common increasing subsequence of both sequences.

 

Sample Input

1

5

1 4 2 5 -12

4

-12 1 2 4

 

Sample Output

2

 

Source

ACM暑期集训队练习赛（二）

 

Recommend

lcy

此题是最长上升子序列的升级版，就是求两个序列的相同最长上升子序列，思路和简单的最长上升子序列相似，只不过多了一个判断第二个序列是否合适的条件，第二个序列存储的时候用结构体记下位置和数值，然后排序，在第一个序列里面求最长上升子序列，状态转移方程如下：

c[i]=max(c[j]+1)(1<=j<i,a[j]<a[i]并且在第二个序列中a[j]，a[i]的位置都存在且a[j]的位置小于a[i]的位置)；还要注意一点，题目没说清楚两个测试数据之间要输出一个空行的，害我PE了两次，郁闷啊

代码如下：

#include<stdio.h>

#include<stdlib.h>

struct point {

int num,index;

}b[505];

int a[505],c[505];

int cmp(const void *e,const void *f)

{

struct point *c,*d;

c=(struct point *)e;

d=(struct point *)f;

return c->num-d->num;

}

int search(int snum,int n)

{

int low,hig,mid;

low=0;hig=n-1;

while(low<=hig)

{

mid=(low+hig)/2;

if(snum>b[mid].num)

low=mid+1;

else if(snum==b[mid].num)

return b[mid].index;

else hig=mid-1;

}

if(b[mid].num==snum)return b[mid].index;

return -1;

}

int main()

{

int i,j,m,n,t,s,k,max,flag=0;

scanf("%d",&t);

for(flag=1;flag<=t;flag++)

{

scanf("%d",&m);

for(i=1;i<=m;i++)

scanf("%d",&a[i]);

scanf("%d",&n);

for(i=0;i<n;i++)

{

scanf("%d",&b[i].num);

b[i].index=i;

}

qsort(b,n,sizeof(b[0]),cmp);

if(search(a[1],n)!=-1)

c[1]=1;

else c[1]=0;

for(i=2;i<=m;i++)

{

k=search(a[i],n);

if(k==-1){c[i]=0;continue;}

max=0;

for(j=1;j<i;j++)

if(a[j]<a[i]&&max<c[j])

{

s=search(a[j],n);

if(s<k&&s!=-1)

max=c[j];

}

c[i]=max+1;

}

max=0;

for(i=1;i<=m;i++)

if(max<c[i])

max=c[i];

printf("%d\n",max);

if(flag<t)printf("\n");

}

return 0;

}