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
学了新的模板LICS 最大公共递增子序列
领悟了一天才大概懂了 妈蛋
代码：
#include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int MAXN=550; int a[MAXN]; int b[MAXN]; int t; int al; int bl; int dp[MAXN]; int main(void){ scanf("%d",&t); while(t--){ memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); scanf("%d",&al); for(int i=1;i<=al;i++){ scanf("%d",&a[i]); } scanf("%d",&bl); for(int i=1;i<=bl;i++){ scanf("%d",&b[i]); } memset(dp,0,sizeof(dp)); int ans=0; for(int i=1;i<=al;i++){ int len=0; for(int j=1;j<=bl;j++){ //这部分更新dp[j] //因为是公共子序列所以只要用b[j]来保存这个公共子序列的最后一个元素 //所以这里a[i]和b[j]比较其实也是和整个公共子序列比较，如果是递增 //则暂时增加len，然后接下来的b[j]循环中如果出现相同的，该LICS就成立，dp就等于len+1 if(a[i]>b[j]){ len=max(len,dp[j]); } //如果两个数相同，说明在前一步的len的基础上再加1，用max保证更优 else if(a[i]==b[j]){ dp[j]=max(dp[j],len+1); } //更新答案 ans=max(ans,dp[j]); } } printf("%d\n",ans); if(t!=0){ printf("\n"); } } return 0; }