10534

最新推荐文章于 2024-07-29 15:27:50 发布

转载最新推荐文章于 2024-07-29 15:27:50 发布 · 138 阅读

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原文链接：http://www.cnblogs.com/ACKOKO/articles/2119922.html

本文介绍了一种使用两遍O(nlogn)复杂度求解最长上升子序列问题的算法实现，通过具体代码展示了如何计算正序及逆序数组的最长上升子序列长度，并结合两者得出最终结果。

要用两遍O（nlogn）的最长上升子序列。

//============================================================================
// Name        : 10534.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <cmath>
using namespace std;

int n, Max, t, top;
int a[10010], l[10010], r[10010], b[10010], stack[10010];



int main() {
	freopen("a.txt", "r", stdin);
	while(scanf("%d", &n)!=EOF){
		for(int i = 1;i <= n;i++){
			scanf("%d", &a[i]);
			b[n+1-i] = a[i];
			l[i] = 1;
			r[i] = 1;
		}
		top = 1;
		stack[top] = a[1];
		l[1] = 1;
		for(int i = 2;i <= n;i++){
			if(a[i] > stack[top]){
				top++;
				stack[top] = a[i];
				l[i] = top;
			}
			else{
				int tag = 0;
				int le = 1, ri = top, mid;
				while(le <= ri){
					mid = floor((le+ri)/2);
					if(stack[mid] == a[i]){
						break;
					}
					if(a[i] >= stack[mid]&&a[i] <= stack[mid+1]){
						stack[mid+1] = a[i];
						break;
					}
					if(le == ri){
						if(a[i] < stack[le]){
							stack[le] = a[i];
							break;
						}
					}
					if(a[i] > stack[mid]){
						le = mid;
					}
					else{
						ri = mid;
					}
				}
				l[i] = top;
			}
		}
		top = 1;
		stack[top] = b[1];
		r[1] = 1;
		for(int i = 2;i <= n;i++){
			if(b[i] > stack[top]){
				top++;
				stack[top] = b[i];
				r[i] = top;
			}
			else{
				int tag = 0;
				int le = 1, ri = top, mid;
				while(le <= ri){
					mid = floor((le+ri)/2);
					if(stack[mid] == b[i]){
						break;
					}
					if(b[i] >= stack[mid]&&b[i] <= stack[mid+1]){
						stack[mid+1] = b[i];
						break;
					}
					if(le == ri){
						if(b[i] < stack[le]){
							stack[le] = b[i];
							break;
						}
					}
					if(b[i] > stack[mid]){
						le = mid;
					}
					else{
						ri = mid;
					}
				}
				r[i] = top;
			}
		}
		Max = 0;
		for(int i = 1;i <= n;i++){
//			printf("%d %d\n\n", l[i], r[i]);
			t = min(l[i], r[n+1-i]);
//			printf("%d\n", t);
			if(Max < t*2-1){
				Max = t*2-1;
			}
		}
		printf("%d\n", Max);
	}
	return 0;
}

转载于:https://www.cnblogs.com/ACKOKO/articles/2119922.html