二分图最大匹配
#962
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优先从左边集合的未匹配点找增广路的话,就只需要以左边集合的未匹配点为起点找了。假如以左边集合的未匹配点为起点找不到增广路,此时以右边集合的未匹配点为起点找找到了一条增广路,由于增广路的性质决定了它的长度一定为奇数,那么说明这条增广路的终点一定在左边,这和前面的假设相违背。 |
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请问为什么没有提到 |
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求增广路代码的注释 |
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请问这段代码中, augment_path(int _n, int _m) : n(_n), m(_m) {
assert(0 <= n && 0 <= m);
pa = vector<int>(n, -1);
pb = vector<int>(m, -1);
vis = vector<int>(n);
g.resize(n);
res = 0;
dfn = 0;
} |
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https://oi-wiki.org/topic/graph-matching/bigraph-match/
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