weixin_45113721的博客

https://www.acwing.com/problem/content/2196/#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int N=5010,M=200010,inf=1e8;int S,T;int h[N],e[M],f[M],w[M],ne[M],idx;int q[N],d[N]

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int N=5010,M=200010,inf=1e8;int S,T;int h[N],e[M],f[M],w[M],ne[M],idx;int q[N],d[N],pre[N],incf[N];bool st[N];void add(int a,i

最大流：#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>using namespace std;const int N=210,M=(10300+210)*2,inf=1e8;int h[N],edge[M],f1[M],f2[M],ver[M],ne[M],tot;int num[N];int s,t;i

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>using namespace std;const int N=210,M=(10300+210)*2,inf=1e8;int h[N],edge[M],f1[M],f2[M],ver[M],ne[M],tot;int num[N];int s,t;int d[

题目：https://www.acwing.com/problem/content/2177/#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>using namespace std;const int N=200+10,M=(110*110+210)*2,inf=1e8;int h[N],ver[M],edge[

题目：https://ac.nowcoder.com/acm/contest/5666/H思路：有1e5次询问，如果对于每次询问都用一次最小费用最大流肯定会超时1.以容量为1建图求最小费用最大流，所求出的每条增广路的流量都是1(因为所有边的容量都是1)。因为所有边的容量都是一样的而且所有边的花费不变，所以所有询问的增广路应该是一样的。2.对于每个询问容量为u/v,流量为1：cost(u/v,1)=cost(u,v)/v（容量和流量同时改变一个倍数，所选择的的增广路不变）3. 每条边的容量变为u,

最大流1.FF方法2.EK算法时间复杂度：O(n*m^2)实际运用中远远达不到这个上界，一般能处理10^3 - 10^4规模的网络#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>using namespace std;const int inf=1<<29,N=2010,M=20010;int