"Codeforces Round #493 (Div. 2) B" cutting

"Link" I am the link, point me:)
Test instructions

Enter the test instructions here

Exercises

Obviously only in the first I position odd even occurrences of the same place can be cut.
(and no matter how the front is cut, here can be cut.)
Then the equivalent of n items, each item cost is |a[i]-a[i-1]|, then the value is 1.
Ask you what is the maximum value of the cost of not exceeding B.
This translates into a 01 backpack model.
Or you can write it that way.
\ (Dp[i][j] Indicates the first I item selected (cut) the minimum cost of J)
Then Dp[i][j]=min (Dp[i-1][j-1]+cost[i],dp[i-1][j]), or//For each of the cut I and not cut the first I.
Then traverse the dp[n][i from large to small]; the output is the largest so that the dp[n][i]<=b I can

Code

#include <bits/stdc++.h>using namespace Std;const int N = 100;const int INF = 1e8;//dp[i][j] First I position, cut the minimum cost of J times//dp[i ][j] = min (dp[i-1][j-1]+cost[i],dp[i-1][j]); int Dp[n+10][n+10],n,b;int Pre[n+10][2],a[n+10],cost[n+10],cnt;int Main    () {#ifdef local_define freopen ("Rush.txt", "R", stdin);    #endif//Local_define ios::sync_with_stdio (0), Cin.tie (0);    CIN >> N >> B;    for (int i = 1;i <= n;i++) cin >> A[i];        for (int i = 1;i <= n;i++) {for (int j = 0;j < 2;j++) Pre[i][j] = Pre[i-1][j];    pre[i][a[i]&1]++;        } for (int i = 1;i <= n-1;i++) if (pre[i][0]==pre[i][1]) {cost[++cnt] = ABS (A[i+1]-a[i]);    } n = cnt;    for (int i = 0;i <= n;i++) for (int j = 0;j <= n;j++) dp[i][j] = INF;    Dp[0][0] = 0;            for (int i = 1;i <= n;i++) for (int j = 0;j <= i;j++) {Dp[i][j] = dp[i-1][j];//no cut I if (j>0) {dp[i][j] = min (dp[i][J],dp[i-1][j-1]+cost[i]);//cut i}} for (int j = n;j >= 0;j--) {if (dp[n][j]<=b) {            cout<<j<<endl;        return 0; }} return 0;}

"Codeforces Round #493 (Div. 2) B" cutting