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  • UVA 12821 Double Shortest Paths

                      Double Shortest Paths
    Alice and Bob are walking in an ancient maze with a lot of caves and one-way passages connecting
    them. They want to go from cave 1 to cave n. All the passages are difficult to pass. Passages are too
    small for two people to walk through simultaneously, and crossing a passage can make it even more
    difficult to pass for the next person. We define di as the difficulty of crossing passage i for the first time,
    and ai as the additional difficulty for the second time (e.g. the second person’s difficulty is di + ai).
    Your task is to find two (possibly identical) routes for Alice and Bob, so that their total difficulty
    is minimized.
    For example, in figure 1, the best solution is 1 → 2 → 4 for both Alice and Bob, but in figure 2, it’s
    better to use 1 → 2 → 4 for Alice and 1 → 3 → 4 for Bob. It’s always possible to reach cave n
    from cave 1.
    Input
    There will be at most 200 test cases. Each case begins with two integers n, m (1 ≤ n ≤ 500, 1 ≤
    m ≤ 2000), the number of caves and passages. Each of the following m lines contains four integers u,
    v, di and ai (1 ≤ u, v ≤ n, 1 ≤ di ≤ 1000, 0 ≤ ai ≤ 1000). Note that there can be multiple passages
    connecting the same pair of caves, and even passages connecting a cave and itself.
    Output
    For each test case, print the case number and the minimal total difficulty.


    Sample Input

    4 4
    1 2 5 1
    2 4 6 0
    1 3 4 0
    3 4 9 1
    4 4
    1 2 5 10
    2 4 6 10
    1 3 4 10
    3 4 9 10


    Sample Output
    Case 1: 23
    Case 2: 24

    这是偶做的一条MCMF。

    一开始想dij一次最短路。然后用dfs再记录下来。然后换路。在dij一次。

    然后发现这样有可能是错的 。。  因为用一种情况是 2 次都经过了 最短路的一部分 。

    。 是有可能小于走完整条最短路 ,然后再多走一次最短路的。。。

    例如 。。

    后来州神给出来一个样例。

    然而, u ,v 构一条费用为 w 流量为1 。一条为w + a ,流量也为1的边

    然后源和汇到1 , n的费用为0,流量为2 , 求出来就是才是真正的答案 ...

    only strive for your goal , can you make your dream come true ?
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  • 原文地址:https://www.cnblogs.com/hlmark/p/4058431.html
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