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果断1A!!DP:将点和边都视为点，第i个点为奇数时是点，偶数时是边。 设f[i][j]表示左边的人在第i个点上，右边的人在第j个点上时，所花费的最小距离。 接下来就可以用dijkstra+heap来解决 Followed by: Post your reply here: |
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