最长递增子序列

参考博客：

http://qiemengdao.iteye.com/blog/1660229

 

/*  最长递增子序列  LIS */
#include <iostream>
using namespace std;
#include <cstdio>
#include <cstdlib>

int main(){
    int s[100], len[100], n;
    // len[i] 表示到第i个元素为止，最长递增子序列的个数
    while(scanf("%d",&n) != EOF){
         for(int i = 0;i < n;i++){ 
             scanf("%d",&s[i]);
             len[i] = 1;
         }/*  dp思想   O(n*n) */                   
         for(int i = 1;i < n;i++){
             for(int j = i-1;j >= 0;j--){
                 if(s[i] > s[j] && len[i] < (len[j]+1))
                 len[i] = len[j] + 1;
             }
         }
         int maxlen = 1;
         for(int i = 0;i < n;i++){
             cout << len[i] << " ";    
             if(len[i] > maxlen){
                   maxlen = len[i];
             }
         }
         cout << endl;
         cout  << maxlen << endl; 
    }
    system("pause");
}

 

#include <iostream>
using namespace std;
#include <cstdlib>
#include <cstdio>
/*  LIS 可以转化成最长公共子序列问题 见：qiemengdao.iteye.com 详见解释  */ 
int find(int t[], int low, int high, int x){
    while(low <= high){
         int mid = low + (high - low)/2 ;
         if(t[mid] == x)
         return mid;
         else if(t[mid] > x){
               low = mid + 1;     
         }
         else if(t[mid] < x){
             high = mid - 1;     
         }
    }
    return low;
}

int main(){
    /*   LIS 问题 */
    int s[100], end[100];
    int n, k;
    /*  end[i] 表示当前递增子序列有i个¸以end[i]结尾，end[i](s[j]中的某个元素)  */
    /* dp 思想 */
    while(scanf("%d",&n) != EOF){
         for(int i = 0;i < n;i++){
            scanf("%d",&s[i]);
         }
         end[0] = s[0];
         k = 1;
         for(int i = 1;i < n;i++){
            if(s[i] >= end[k-1]){
               end[k++] = s[i];
            }
            else{
                 int pos = find(end,0,k-1,s[i]);
                 end[pos] = s[i];
            }
         }
         for(int i = 0;i < k;i++){
            cout << end[i] << " ";        
         }
         cout << endl;
         cout << k << endl;
    }
    system("pause");
}

具体应用见：

http://hawstein.com/posts/9.7.html  叠罗汉