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
题意：求最长递增公共子序列的长度
思路：直接模板
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std; int n,m,a[505],b[505],dp[505][505]; int LICS()
{
int MAX,i,j;
memset(dp,0,sizeof(dp));
for(i = 1; i<=n; i++)
{
MAX = 0;
for(j = 1; j<=m; j++)
{
dp[i][j] = dp[i-1][j];
if(a[i]>b[j] && MAX<dp[i-1][j])
MAX = dp[i-1][j];
if(a[i]==b[j])
dp[i][j] = MAX+1;
}
}
MAX = 0;
for(i = 1; i<=m; i++)
if(MAX<dp[n][i])
MAX = dp[n][i];
return MAX;
} int main()
{
int i,t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i = 1; i<=n; i++)
scanf("%d",&a[i]);
scanf("%d",&m);
for(i = 1; i<=m; i++)
scanf("%d",&b[i]);
printf("%d\n",LICS());
if(t)
printf("\n");
} return 0;
}
上面的虽然可以解决，但是二维浪费空间较大，我们注意到在LICS函数中有一句dp[i][j] = dp[i-1][j]，这证明dp数组前后没有变化！于是可以优化成一维数组！
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; int a[505],b[505],dp[505],n,m; int LICS()
{
int i,j,MAX;
memset(dp,0,sizeof(dp));
for(i = 1; i<=n; i++)
{
MAX = 0;
for(j = 1; j<=m; j++)
{
if(a[i]>b[j] && MAX<dp[j])
MAX = dp[j];
if(a[i]==b[j])
dp[j] = MAX+1;
}
}
MAX = 0;
for(i = 1; i<=m; i++)
if(MAX<dp[i])
MAX = dp[i];
return MAX;
} int main()
{
int t,i;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i = 1; i<=n; i++)
scanf("%d",&a[i]);
scanf("%d",&m);
for(i = 1; i<=m; i++)
scanf("%d",&b[i]);
printf("%d\n",LICS());
if(t)
printf("\n");
} return 0;
}