Array Without Local Maximums CF-1068D（计数DP）

题意：

给一个序列$a$，对于每一位$a[i]$，要求满足$a[i]<=a[i-1]$或者$a[i]<=a[i+1]$，如果$a[i]=-1$，则表示该位未知，问最多有多少种序列可以满足条件。

思路：

$dp[i][j][k]$表示第$i$位填充$j$的时候与前一位的关系是$k$。

$k=0$表示$a[i]>a[i-1]$，$k=1$表示$a[i]=a[i-1]$，$k=2$表示$a[i]<a[i-1]$。转移式为：

$dp[i][j][0]=\sum_{k=1}^{j-1}{(dp[i-1][k][0]+dp[i-1][k][1]+dp[i-1][k][2])}$

$dp[i][j][1]=dp[i-1][j][0]+dp[i-1][j][1]+dp[i-1][j][2]$

$dp[i][j][2]=\sum_{k=j+1}^{200}{(dp[i-1][k][1]+dp[i-1][k][2])}$

$ans=\sum_{j=1}^{200}{(dp[n][j][1]+dp[n][j][2])}$

代码：

 1 //#include<bits/stdc++.h>
 2 #include <set>
 3 #include <map>
 4 #include <stack>
 5 #include <cmath>
 6 #include <queue>
 7 #include <cstdio>
 8 #include <string>
 9 #include <vector>
10 #include <cstring>
11 #include <iostream>
12 #include <algorithm>
13 
14 #define ll long long
15 #define pll pair<ll,ll>
16 #define pii pair<int,int>
17 #define bug printf("*********\n")
18 #define FIN freopen("input.txt","r",stdin);
19 #define FON freopen("output.txt","w+",stdout);
20 #define IO ios::sync_with_stdio(false),cin.tie(0)
21 #define ls root<<1
22 #define rs root<<1|1
23 #define pb push_back
24 
25 using namespace std;
26 const int inf = 2e9 + 7;
27 const ll Inf = 1e18 + 7;
28 const int maxn = 1e5 + 5;
29 const ll mod = 998244353;
30 
31 ll dp[maxn][201][3];
32 int a[maxn];
33 
34 int main()
35 {
36     int n;
37     scanf("%d", &n);
38     for (int i = 1; i <= n; ++i)    scanf("%d", &a[i]);
39     if (a[1] == -1) {
40         for (int i = 1; i <= 200; ++i)    dp[1][i][0] = 1;
41     }
42     else {
43         dp[1][a[1]][0] = 1;
44     }
45     for (int i = 2; i <= n; ++i)
46     {
47         ll tmp = 0;//大于
48         for (int j = 1; j <= 200; ++j) {
49             if (a[i] == -1 || a[i] == j)    dp[i][j][0] = tmp;
50             else dp[i][j][0] = 0;
51             tmp += dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][2];
52             tmp %= mod;
53         }
54         //等于
55         for (int j = 1; j <= 200; ++j) {
56             if (a[i] == -1 || a[i] == j)
57                 dp[i][j][1] = (dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][2]) % mod;
58             else dp[i][j][1] = 0;
59         }
60         tmp = 0;//小于
61         for (int j = 200; j >= 1; --j) {
62             if (a[i] == -1 || a[i] == j)    dp[i][j][2] = tmp;
63             else dp[i][j][2] = 0;
64             tmp += dp[i - 1][j][1] + dp[i - 1][j][2];
65             tmp %= mod;
66         }
67     }
68     ll ans = 0;
69     for (int i = 1; i <= 200; ++i) {
70         if (a[n] == -1 || a[n] == i) {
71             ans += dp[n][i][1] + dp[n][i][2];
72             ans %= mod;
73         }
74     }
75     printf("%lld\n", ans);
76 }

标签：tmp,Maximums,sum,CF,1068D,long,include,dp,define
来源： https://www.cnblogs.com/zhang-Kelly/p/12719138.html