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题意:动态修改图 \(G\) 的边集,求每次修改后的 \(\sum c\times (c−1) / 2\) (记每个强连通分量中的点数量为 \(c\) ).其中修改操作共 \(m\) 次,每次最多改 \(k\) 条边.\(1≤m≤25000,1≤k≤10\),图 \(G\) 中点数为 \(n\) ,\(1≤n≤250\). 题解:\(tarjan\)由于要遍历所有边,复杂度 \(\mathcal{O}((V+E)\times m)\),在稠密图中效率较低:而使用邻接矩阵存储边集状态,利用 \(K…
Kosaraju算法,然後bitset優化 主要是學習一下自寫bitset的姿勢 #include<cstring> #include<algorithm> #include<cstdio> #include<cmath> #define rep(i,l,r) for(int i=l;i<=r;i++) #define dow(i,l,r) for(int i=r;i>=l;i--) #define rep0(i,r) for(int i=0;i…
http://acm.hdu.edu.cn/showproblem.php?pid=6072 题意: 给你$n*n$的矩阵,每次修改k条边,让你计算其中能相互到达的点对有多少. 思路: 其实就是求强连通分量,如果一个强连通分量里有n个点,那么这里面的点对就有$n*(n-1)/2$.用Kosaraju计算即可,但是这样是会超时的,还需要用bitset来优化. __builtin_函数中处理二进制位的函数: int __builtin_ffs (unsigned x) 返回x中最后一个1是从右往左第…
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