CodeForces 447C DZY Loves Sequences DP

题目：click here

题意：求给定序列更改其中一个元素后的最长连续上升子序列的长度

分析：最长的连续子序列有2种，一种是严格上升(没有更改元素)的长度加1，一种是两段严格上升的加起来。、

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define F first
 4 #define S second
 5 #define pb push_back
 6 #define power(a) ((a)*(a))
 7 #define ENTER printf("\n");
 8 typedef unsigned long long ll;
 9 const int INF = 0x3f3f3f3f;
10 const double EPS = 1e-8;
11 const int M = 1e5+5;
12
13 int n;
14 int a[M];
15 int dps[M];     // dps[i]保存以i开头的最长连续上升子序列
16 int dpe[M];     // dpe[i]保存以i结束的最长连续上升子序列
17 void solve()    {
18     dpe[0] = 1;
19     dps[n-1] = 1;
20     int ret = 1;
21     for( int i=1; i<n; i++ ) {
22         if( a[i] > a[i-1] ) dpe[i] = dpe[i-1] + 1;
23         else dpe[i] = 1;
24         ret = max( ret, dps[i] );
25     }
26     for( int i=n-2; i>=0; i-- ) {
27         if( a[i] < a[i+1] ) dps[i] = dps[i+1] + 1;
28         else dps[i] = 1;
29         ret = max( ret, dps[i] );
30     }
31     for( int i=0; i<n; i++ )  {
32         if( i == 0 )
33             ret = max( ret, dps[i]+1 );
34         else if( i == n-1 )
35             ret = max( ret, dpe[i]+1 );
36         else    {
37             if( a[i+1]-a[i-1] >= 2 )
38                 ret = max( ret, dps[i+1]+dpe[i-1]+1 );
39             else
40                 ret = max( ret, max( dps[i+1]+1, dpe[i-1]+1 ) );
41         }
42     }
43     printf("%d\n", ret>n?n:ret );   // 结果最大为n
44 }
45 int main()  {
46     while( ~scanf("%d", &n ) )  {
47         for( int i=0; i<n; i++ )
48             scanf("%d", a+i);
49         solve();
50     }
51     return 0;
52 }