Luo Valley P1880 [NOI1995] combined stone (section DP)

Ok...

 

Topic link: https: //www.luogu.org/problem/P1880

 

This question is characterized by a gravel ring, so let a [i + n] = a [i] (twice the length) of the ring to solve the problem, then note the minimum time interval required to be initialized to a large dp the number ...

 

AC Code:

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 
 6 using namespace std;
 7 
 8 int dp1[205][205], dp2[205][205], sum[205], a[205];
 9 
10 int main(){
11     int n;
12     scanf("%d", &n);
13     for(int i = 1; i <= n; i++) {scanf("%d", &a[i]); a[i + n] = a[i];}
14     for(int i = 1; i <= n * 2; i++) {sum[i] = sum[i - 1] + a[i];}
15     for(int l = 1; l <= n; l++){
16         for(int i = 1; i <= 2 * n; i++){
17             int j = i + l;
18             dp2[i][j] = 0x3f3f3f;//Initialized for the minimum lot 
. 19              IF (J> 2 * n-) BREAK ;
 20 is              for ( int K = I; K <J; K ++ ) {
 21 is                  DP1 [I] [J] = max (DP1 [I] [J ], DP1 [I] [K] + DP1 [K + . 1 ] [J] + SUM [J] - SUM [I - . 1 ]);
 22 is                  DP2 is [I] [J] = min (DP2 is [I] [J ], DP2 is [I] [K] + DP2 is [K + . 1 ] [J] + SUM [J] - SUM [I - . 1 ]);
 23 is              }
 24          }
 25      }
 26 is      int Maxx = 0 , Minn = 0x3f3f ;
 27     for(int i = 1; i <= n; i++){
28         int j = i + n - 1;
29         maxx = max(maxx, dp1[i][j]);
30         minn = min(minn, dp2[i][j]);
31     }
32     printf("%d\n%d\n", minn, maxx);
33     return 0;
34 }

AC Code