poj 3670 Eating Together

//最长递增子序列（LIS）和最长递减子序列（LDS）的解法其实是一样的，题目就是要求需要改动的次数最少，所以只要
//求出其序列的最长递增子序列和最长递减子序列，通过比较就可以最长的子序列，从而就可以求出改动次数最少答案了！ 
#include <iostream>
#include <cstdio> 
#include <memory.h> 
#include <climits> 
#include <algorithm> 
using namespace std; 
const int MAX = 30001; 
int num[MAX], tmp[MAX];

int main()
{
    int i, n, ans, ans1, ans2, left, right, mid;
    scanf("%d", &n);
    for (i = 0; i < n; i++){
        scanf("%d", &num[i]); 
    } 
    
    //最长递增子序列的求解 
    memset(tmp, 0, sizeof(tmp)); 
    tmp[0] = -1;
    ans1 = 0;
    for (i = 0; i < n; i++){
        //这里是要判断是否有相等的，因为题目只是给出了1到3的数，而它的递增序列可以是112233这样的形式！ 
        if (num[i] >= tmp[ans1]){
            tmp[++ans1] = num[i]; 
        } 
        else {
             left = 1, right = ans1;
             while (left <= right){
                   mid = (left + right) / 2;
                   if (num[i] >= tmp[mid]){
                       left = mid + 1; 
                   } 
                   else{
                        right = mid - 1; 
                   } 
             } 
             tmp[left] = num[i]; 
        } 
    }
     
     
    //最长递减子序列的求解 
    memset(tmp, 0, sizeof(tmp)); 
    tmp[0] = INT_MAX;
    ans2 = 0; 
    for (i = 0; i < n; i++){
        if (num[i] <= tmp[ans2]){
            tmp[++ans2] = num[i]; 
        } 
        else {
             left = 1, right = ans2;
             while (left <= right){
                   mid = (left + right) / 2;
                   if (num[i] <= tmp[mid]){
                       left = mid + 1; 
                   } 
                   else{
                        right = mid - 1; 
                   } 
             } 
             tmp[left] = num[i]; 
        } 
    }
    
    //cout << ans1 << " " << ans2 << endl; 
    ans = n - max(ans1, ans2);
    cout << ans << endl; 
    
    system("pause"); 
} 
 
 
/* 
5
1
1
2
3
1
*/