uva-10534 Wavio Sequence

标签：

题目链接

题目大意

       从前往后n + 1个数都是递增的，然后从递增的最大一个值达到最后都是递减的也是n个数总共2*n+1个数。给你一个值求最长的一个满足上面要求的序列。

思路：

       从前往后求一次单调递增子序列，从后往前求一次单调递增子序列，每次记录下值然后记录最大值

代码：

#include<iostream>
#include<cstdio>
#include<map>
#include<math.h>
#include<cstring>
#include<algorithm>
#define INF 0x1f1f1f1f
using namespace std;
int main()
{
    int n, i, a[10010];
    while(scanf("%d",&n) != EOF)
    {
        for(i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
        }
        int t1 = 0, t2 = 0;
        int dp1[10010], s[10010], dp2[10010];
        memset(dp1, 0, sizeof(dp1));
        memset(s, -1, sizeof(s));
        int top = 0;
        s[0] = -INF;
        //从前往后求出单调递增子序列
        for(i = 1; i <= n; i++)
        {
            if(a[i] > s[top])
            {
                s[++top] = a[i];
                dp1[i] = top;
            }
            else
            {
                int l = 1, r = top;
                while(l <= r)
                {
                    int mid = (l + r) / 2;
                    if(a[i] > s[mid])
                    {
                        l = mid + 1;
                    }
                    else
                        r = mid - 1;
                }
                s[l] = a[i];
                dp1[i] = l;
            }
            if(dp1[i] > t1)
                t1 = dp1[i];
        }
        memset(dp2, 0, sizeof(dp2));
        memset(s, -1, sizeof(s));
        top = 0;
        s[0] = -INF;
        //从后往前求出单调递增子序列
        for(i = n; i > 0; i--)
        {
            if(a[i] > s[top])
            {
                s[++top] = a[i];
                dp2[i] = top;
            }
            else
            {
                int l = 1, r = top;
                while(l <= r)
                {
                    int mid = (l + r) / 2;
                    if(a[i] > s[mid])
                    {
                        l = mid + 1;
                    }
                    else
                        r = mid - 1;
                }
                s[l] = a[i];
                dp2[i] = l;
            }
            if(dp2[i] > t2)
                t2 = dp2[i];
        }
        int ans = 1,tem;
        //每次求最小值
        for(int i = 1;i <= n;i++)
        {
            tem = min(dp1[i],dp2[i]);
            ans = max(tem*2-1,ans);
        }
        cout << ans << endl;

    }
    return 0;
}

uva-10534 Wavio Sequence

标签：

原文地址：http://blog.csdn.net/bmamb/article/details/51333262