dp 最大递增子序列

Longest Ordered Subsequence

Time Limit: 2 Seconds Memory Limit: 65536 KB

A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2
< ... < iK <= N. For example, the sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e.g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences of this sequence are of length 4, e.g., (1, 3, 5, 8).
Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

Input
The first line of input contains the length of sequence N (1 <= N <= 1000). The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated
by spaces.

Output
Output must contain a single integer - the length of the longest ordered subsequence of the given sequence.

This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank
line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.



Sample Input
1

7

1 7 3 5 9 4 8

Sample Output
4

思想：在a[i]前面找到a[j] 　<　a[i](j >= 0 && j < i)的最大b[j],则以a[i]结尾的最长递增子序列的长度为b[j] +１。

#include<stdio.h>
int main()
{
int a[10010],b[10010];//a数组记录数据， b数组记录从第一个数组每个数字的最大单调递增子序列。
int n, i, m, max, j, flag = 0;
scanf("%d",&n);
while(n--)
{
scanf("%d",&m);
for(i = 0; i < m; i++)
scanf("%d",&a[i]);
max = 1;
b[0] = 1;
for(i = 1; i < m; i++)
{
int k = 0;
for(j = 0; j < i; j++)
if(a[j] < a[i] && k < b[j])//找到满足a[j] < a[i] 的最大b[j];
{
k = b[j];
}
b[i] = k + 1;
if(max < b[i])
max = b[i];
}
if(flag)
printf("\n");
else
flag = 1;
printf("%d\n",max);
}
return 0;
}

可以当做模板使用哟！！！