一:
很容易想到的 DP的O(N^2)的复杂度
#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
using namespace std;
#define clc(a, b) memset(a, b, sizeof(a))
const int inf = 0x3f;
const int INF = 0x3f3f3f3f;
const int maxn = 1000;
int n, a[maxn], dp[maxn];
int LIS()
{
int i, j, k = 0;
for(i = 0; i < n; i++)
{
dp[i] = 1;
for(j = 0; j < i; j++)
{
if(a[i] > a[j] && dp[i] < dp[j] + 1)
{
dp[i] = dp[j] + 1;
}
}
k = dp[i] > k ? dp[i] : k;
}
return k;
}
int main()
{
while(~scanf("%d", &n))
{
clc(dp, 0);
for(int i = 0; i < n; i++)
scanf("%d", &a[i]);
printf("LIS = ");
printf("%d\n", LIS());
}
}
二: 扩展升级版, 求定长的上升子序列个数
原文:http://www.cnblogs.com/tenlee/p/4539494.html