OJ-Problems-Source/.ACM-Templates/LIS.cpp
2016-08-11 09:05:46 +08:00

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#include <stdio.h>
#define MAXN 1010
int seq[MAXN];
int seqlen[MAXN];
int main()
{
int N;
while(scanf("%d",&N)==1)
{
int i,j,k,max,maxlen=1;
for(i=1; i<=N; i++)
seqlen[i]=1; //seqlen数组存以第i个数为终点的最长上升序列
for(i=1; i<=N; i++)
scanf("%d",&seq[i]); //seq数组保存序列数组
for (i=2; i<=N; i++)
{
max=0;
for (j=1; j<=i-1; j++)
{
if(seq[j]<seq[i]&&seqlen[j]>max) //在前i-1个序列中寻找以终点小于seq[i]的最长的子序列,即最优子状态
max=seqlen[j];
}
seqlen[i]=max+1;
if(seqlen[i]>maxlen) //seqlen中保存的是第i个数为终点的最长上升序列找出这个数组中最大的值即为最优序列长度
maxlen=seqlen[i];
}
printf("%d\n",maxlen);
}
return 0;
}