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为了解决这个问题，我们需要计算《三国志》中小白能得到的最大财宝数量。小白需要派出武将攻占其他城池，每走一公里消耗一石粮食。我们需要找到从0号城池出发，到其他各城池的最短路径，然后根据粮食限制，选择能到达的最多财宝的城池。

方法思路

解决代码

import sysimport heapqdef main():    input = sys.stdin.read().split()    ptr = 0    T = int(input[ptr])    ptr += 1    for _ in range(T):        S = int(input[ptr])        N = int(input[ptr+1])        M = int(input[ptr+2])        ptr +=3                # Read M roads        adj = [[] for _ in range(N+1)]        for __ in range(M):            A = int(input[ptr])            B = int(input[ptr+1])            C = int(input[ptr+2])            ptr +=3            adj[A].append((B, C))            adj[B].append((A, C))                # Read V array        V = []        for i in range(N):            V.append(int(input[ptr]))            ptr +=1                # Dijkstra algorithm        INF = float('inf')        d = [INF] * (N+1)        d[0] = 0        heap = []        heapq.heappush(heap, (0, 0))                while heap:            dist_u, u = heapq.heappop(heap)            if dist_u > d[u]:                continue            for v, w in adj[u]:                if d[v] > d[u] + w:                    d[v] = d[u] + w                    heapq.heappush(heap, (d[v], v))                # Collect available cities        cities = []        for i in range(1, N+1):            if d[i] <= S:                cities.append((d[i], V[i-1]))  # V is 0-based                # 0-1 Knapsack        max_S = S        dp = [0] * (max_S + 1)        for (weight, value) in cities:            for j in range(max_S, weight-1, -1):                if dp[j - weight] + value > dp[j]:                    dp[j] = dp[j - weight] + value                print(dp[max_S])if __name__ == '__main__':    main()

代码解释

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