Codeforces 1497D DP

Title

Portal Codeforces 1497D Genius

answer

Observe $|c_i-c_j|$，设 $i<j$ , the result is the binary representation$[i,j-1]$ bit is$1$ number; then for different$(i,j)$ 对，$|c_i-c_j|The$ value is different. If the state is expressed as a two-tuple composed of the end point of the path and the maximum value that can be taken on the upper edge of the path, use$|c_i-c_j|Connect$ the edges for the edge rights, then get a$DAG$ , and the edge weight on the path increases monotonically.

$dp\left[i\right]\left[k\right]$ represents the path with$i$ is the end point, and the maximum value that can be taken on the road strength is$When\; k$ , the maximum number of points that can be obtained. Let$k=|c_i-c_j|$，且 $ta{g}_i\ne ta{g}_j$设 设$pre\left(k\right)$ is less than$k$ 的最大边权，则有
{ d p [ i ] [ k ] = max ⁡ { d p [ i ] [ p r e ( k ) ] , d p [ j ] [ p r e ( k ) ] + ∣ c i − c j ∣ } d p [ j ] [ k ] = max ⁡ { d p [ j ] [ p r e ( k ) ] , d p [ i ] [ p r e ( k ) ] + ∣ c i − c j ∣ } \begin{cases}dp[i][k]=\max\{dp[i][pre(k)],dp[j][pre(k)]+ \lvert c_i-c_j\rvert\}\\ dp[j][k]=\max\{dp[j][pre(k)],dp[i][pre(k)]+ \lvert c_i-c_j\rvert\}\end{cases} { dp[i][k]=max{ dp[i][pre(k)],dp[j][pre(k)]+∣ci​−cj​∣}dp[j][k]=max{ dp[j][pre(k)],dp[i][pre(k)]+∣ci​−cj​∣}​DP DP in increasing order of the edge weights that can be selected$DP$，即 $i(1\le i\le n)$ 递增，$j(1\le j\le i-1)$ Decrease; realize the compression to one-dimensional$DP$ . Finally, enumerate the end of the path and update the answer.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 5005;
int T, N, tag[maxn], s[maxn];
ll dp[maxn];

int main()
{
    
    
    scanf("%d", &T);
    while (T--)
    {
    
    
        scanf("%d", &N);
        for (int i = 1; i <= N; ++i)
            scanf("%d", tag + i);
        for (int i = 1; i <= N; ++i)
            scanf("%d", s + i);
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= N; ++i)
            for (int j = i - 1; j; --j)
            {
    
    
                if (tag[i] == tag[j])
                    continue;
                ll xi = dp[i], xj = dp[j], d = abs(s[i] - s[j]);
                dp[i] = max(dp[i], xj + d), dp[j] = max(dp[j], xi + d);
            }
        printf("%lld\n", *max_element(dp + 1, dp + N + 1));
    }
    return 0;
}