POJ 2533 Longest Ordered Subsequence （模版LIS）

标签：poj   2533   c++   dp   

题意:输出最长递增子序列的长度

思路：直接裸LIS，

</pre><pre code_snippet_id="648389" snippet_file_name="blog_20150420_4_7842089" class="sio" name="code" style="white-space: pre-wrap; word-wrap: break-word; font-size: 14px; line-height: 26px; background-color: rgb(255, 255, 255);"><pre name="code" class="cpp">#include<cstdio>

const int N = 1001;
int a[N], f[N], d[N]; // d[i]用于记录a[0...i]的最大长度
int bsearch(const int *f, int size, const int &a) {
    int l=0, r=size-1;
    while( l <= r ){
        int mid = (l+r)/2;
        if( a > f[mid-1] && a <= f[mid] ) return mid;// >&&<= 换为: >= && <
        else if( a < f[mid] ) r = mid-1;
        else l = mid+1;
    }
}

int LIS(const int *a, const int &n){
    int i, j, size = 1;
    f[0] = a[0]; d[0] = 1;
    for( i=1; i < n; ++i ){
        if( a[i] <= f[0] ) j = 0; // <= 换为: <
        else if( a[i] > f[size-1] ) j = size++;// > 换为: >=
        else j = bsearch(f, size, a[i]);
        f[j] = a[i]; d[i] = j+1;
    }
    return size;
}

int main()
{
    int i, n;
    while( scanf("%d",&n) != EOF ){
    for( i=0; i < n; ++i ) scanf("%d", &a[i]);
        printf("%d\n", LIS(a, n)); // 求最大递增/上升子序列(如果为最大非降子序列,只需把上面的注释部分给与替换)
    }
    return 0;
}

POJ 2533 Longest Ordered Subsequence （模版LIS）

标签：poj   2533   c++   dp   

原文地址：http://blog.csdn.net/wangxinxin_/article/details/45150899