CodeForces 447C DZY Loves Sequences (dp 子序列)

因为change at most one number (change one number to any integer you want)
所以找到左边递增的最大区间长度

然后找到右边递增的最大区间长度

如果最大长度就是n，那么说明原来就是升序排好了的。

注意以下数据：

5

1 2 2 3 4

输出： 4

有个细节见下方代码

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<deque>
#include<functional>
#include<iterator>
#include<vector>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#define CPY(A, B) memcpy(A, B, sizeof(A))
typedef long long LL;
typedef unsigned long long uLL;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-9;
const double OO = 1e20;
const double PI = acos (-1.0);
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
using namespace std;
const int maxn=1e5+10;
struct AA {
int left/*left length*/,right,element;
};
AA dp[maxn];
int main() {
int n; scanf ("%d",&n);
memset (dp,0,sizeof (dp) );
for (int i=1; i<=n; i++) {
scanf ("%d",&dp[i].element);
}
if (n==1) {printf ("1\n");}/**/
else {
int ans=0;
dp[1].left=1;
for (int i=2; i<=n; i++) {
dp[i].left= (dp[i].element>dp[i-1].element?dp[i-1].left+1/*length add 1*/:1);
ans=max (ans,dp[i].left);//大于左边的，那么就是左边长度再加1
}
dp
.right=1;
for (int i=n-1; i>=1; i--) {
dp[i].right= (dp[i+1].element>dp[i].element?dp[i+1].right+1:1);
ans=max (ans,dp[i].right);
}
if (ans==n) { printf ("%d",n); }//等于n意味着之前顺序就是排好的
else {
for (int i=1; i<=n; i++) {
if (dp[i+1].element>dp[i-1].element+1) {//must be dp[i+1]>dp[i-1]+1,because: 1 ? 2 and 1 ? 1
ans=max (ans,dp[i-1].left+dp[i+1].right+1);
} else {
ans=max (ans,max (dp[i].left+1,dp[i].right+1) );
}
}
printf ("%d\n",ans);
}
}
return 0;
}