POJ-2533 Longest Ordered Subsequence

http://poj.org/problem?id=2533

该题是最裸的LIS题，这里有两种方法。

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;

int a[10000], dp[10004];

int main(){int N;while (scanf("%d", &N) == 1)    {int ans = 1, m;for (int i = 0; i < N; ++i)        {            scanf("%d", a+i);         }        dp[0] = 1;for (int i = 1; i < N; ++i)        {            m = 0;for (int j = 0; j < i; ++j)            {if (a[i] > a[j] && dp[j] > m)                {                    m = dp[j];                }            }            dp[i] = m+1;            ans = max(ans, dp[i]);        }         printf("%d\n", ans);    } 

}

二分优化后的写法，数组c[i]保留长度为i的上升串的末尾最小元素。

代码如下：

#include <cstdlib>#include <cstring>#include <cstdio>#define MAXN 10000using namespace std;

int a[MAXN+5];

int bsearch(int *c, int l, int r, int val){int mid;while (l <= r)    {        mid = (l+r)>>1;if (val > c[mid])            l = mid + 1;else if (val < c[mid])            r = mid - 1;else  // 如果不是严格的递增序列的话，二分查找时，相等视为后者比其大            return mid;    }return l;}

int LIS(int *a, int N){int c[MAXN], size = 1, pos;    c[1] = a[0];for (int i = 1; i < N; ++i)    {if (a[i]>c[size])            pos = ++size;else            pos = bsearch(c, 1, size, a[i]);// 返回比该值稍微大的点        c[pos] = a[i];    }return size;}

int main(){int N;while (scanf("%d", &N) == 1)    {for (int i = 0; i < N; ++i)        {            scanf("%d", &a[i]);        }        printf("%d\n", LIS(a, N));    }}

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