Overview

This problem asks us to find the minimum physical strength consumed (the sum of costs along the path) to travel from point \(0\) to point \(N+1\) while passing through checkpoints in increasing order of their indices, under the conditions that “the number of visited checkpoints is at least \(K\)” and “the number of visited checkpoints for each terrain type is even”.

Approach

First, notice that the movement always proceeds only in the direction of increasing indices. This can be viewed as a shortest path problem on a DAG (Directed Acyclic Graph) where the points are vertices. We can think of this as having an edge from point \(a\) to point \(b\) (\(a < b\)) with cost \(C_{a,b}\).

However, instead of a simple shortest path, we have the following two additional conditions:

1. Count condition: The number of visited checkpoints \(s \geq K\)
2. Terrain balance condition: For each terrain type \(j\), the number of visited checkpoints of type \(j\) must be even.

We need to find the shortest path while keeping track of these conditions as the “state during the path”. Therefore, we use Dynamic Programming (DP).

Choosing the State

The key is to represent the information required to check the conditions “compactly”.

• Count condition: Although the actual number of visited checkpoints can increase up to \(N\), we only need to know whether it is “at least \(K\)”. Therefore, it is sufficient to cap the count at \(K\) and maintain it as \(\min(\text{count}, K)\). This limits the number of states to \(K+1\) possibilities from \(0 \sim K\).
• Terrain balance condition: Only the parity of the count of each type matters. Since there are \(M\) types (\(M \leq 8\)), we can represent the parity of each type using an \(M\)-bit bitmask. Every time we visit a checkpoint of a certain type, we can flip (XOR) the corresponding bit. Ultimately, if the mask is \(0\) (all types have even counts), the condition is satisfied.

Thus, by capping the count at \(K\) and representing the parities as a bitmask, we can keep the number of states finite and small. This is the core idea of this problem.

Algorithm

We define the DP state as follows:

\[dp[i][c][mask] = \text{現在地点 } i \text{ にいて、立ち寄った個数が } \min(\cdot, K) = c\text{、各タイプの偶奇が } mask \text{ であるときの最小コスト}\]

• Initial state: \(dp[0][0][0] = 0\) (Start at the base, nothing visited yet)
• Transitions: Choose the next checkpoint \(j\) (\(i < j \leq N\)) to visit from the current point \(i\),
  • New count \(nc = \min(c+1, K)\)
  • New mask \(nm = mask \oplus (1 \ll (t_j - 1))\)
  • \(dp[j][nc][nm] \leftarrow \min(dp[j][nc][nm],\ dp[i][c][mask] + C_{i,j})\)
• Termination: Move directly from point \(i\) to the summit \(N+1\). Since the conditions must be satisfied, we consider \(dp[i][c][mask] + C_{i,N+1}\) as a candidate for the answer only when \(c \geq K\) and \(mask = 0\).

Since transitions only occur in the direction of increasing point indices, processing \(i\) in order from \(0\) naturally maintains the topological order of the DP, allowing us to compute it in a single loop.

The minimum value among all candidates is the answer. If no path satisfies the conditions (i.e., there are no candidates), output \(-1\).

Complexity Analysis

The number of states is \(O(N \cdot K \cdot 2^M)\), and there are \(O(N)\) transitions from each state, so:

• Time Complexity: \(O(N^2 \cdot K \cdot 2^M)\) (in the worst case where \(K = N\), \(O(N^3 \cdot 2^M)\))
• Space Complexity: \(O(N \cdot K \cdot 2^M)\)

As the constraints state \(N^3 \cdot 2^M \leq 3 \times 10^8\), this complexity runs sufficiently fast.

Implementation Points

• Capping the count: By setting nc = min(c+1, K), we treat all counts greater than or equal to \(K\) as the same state. If you forget this, the number of states will explode, and you won’t be able to correctly check the “at least \(K\)” condition.

• Bitwise parity flipping: When visiting type \(t_j\), we toggle the parity with mask ^= (1 << (t_j - 1)). Note that since the type indices are 1-based, the bit position will be \(t_j - 1\).

• Flattening multidimensional arrays: In the code, \(dp[i][c][mask]\) is flattened into a \(1\)-dimensional array using the formula idx(i,c,mask) = (i*(K+1)+c)*2^M + mask. This makes memory allocation and access more efficient.

• Handling INF: Since the maximum cost is \(10^9\) and the path length is around \(N+1\), the sum can exceed a \(32\)-bit integer, so we use long long. INF is set to a sufficiently large value (\(4 \times 10^{18}\)) to represent unreachable states.

• The \(s=0\) case: The case of moving directly from the base to the summit without visiting any checkpoints can be naturally handled as a transition from \(dp[0][0][0]\) to \(N+1\) (though this only satisfies the condition when \(K = 0\)).

  Source Code

#include <bits/stdc++.h>
using namespace std;

int main(){
    int N, M, K;
    scanf("%d %d %d", &N, &M, &K);
    vector<int> t(N+1);
    for(int i=1;i<=N;i++) scanf("%d", &t[i]);
    
    static long long C[152][152];
    for(int s=0;s<=N;s++){
        for(int b=s+1;b<=N+1;b++){
            long long x; scanf("%lld",&x);
            C[s][b]=x;
        }
    }
    
    const long long INF = (long long)4e18;
    int MASK = 1<<M;
    // dp[i][c][mask], i in 0..N
    // flatten
    int Kp = K+1;
    auto idx=[&](int i,int c,int mask)->long long{
        return ((long long)i*Kp + c)*MASK + mask;
    };
    long long total = (long long)(N+1)*Kp*MASK;
    vector<long long> dp(total, INF);
    
    dp[idx(0,0,0)] = 0;
    
    long long ans = INF;
    
    for(int i=0;i<=N;i++){
        for(int c=0;c<Kp;c++){
            for(int mask=0;mask<MASK;mask++){
                long long cur = dp[idx(i,c,mask)];
                if(cur==INF) continue;
                // go to N+1
                if(c>=K && mask==0){
                    long long v = cur + C[i][N+1];
                    if(v<ans) ans=v;
                }
                // go to checkpoint j
                for(int j=i+1;j<=N;j++){
                    int nc = min(c+1, K);
                    int nm = mask ^ (1<<(t[j]-1));
                    long long v = cur + C[i][j];
                    long long &ref = dp[idx(j,nc,nm)];
                    if(v<ref) ref=v;
                }
            }
        }
    }
    
    if(ans==INF) printf("-1\n");
    else printf("%lld\n", ans);
    return 0;
}

---

This editorial was generated by claude4.8opus-high.