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"传送门" 有源汇的上下界网络流求最大流……(悄悄嘀咕一句,我也想给小姐姐拍照……) 题目大意:Aya要给一群小姐姐拍照。她在n天(n include include include include include include include include define rep(i,a,n) 阅读全文
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"传送门" 这题是真的恶心…… 有源汇的上下界网络流求可行流…… 首先矩阵的建图基本比较清晰,就是行列之间连边,其上下界由给定的条件决定。这题其实有两种改造法都能过。第一种是最正统的套路,就是首先建立原点和汇点,然后把行向原点连边,容量全都是0(因为上下界的差值是0),不过要更改这些点的流入和流出下 阅读全文
摘要:
"传送门" 题目大意:给定N(N include include include include include include include include define rep(i,a,n) for(int i = a;i = a;i ) define enter putchar('\n') 阅读全文