tarjan算法求解2-SAT问题
题目链接:点击这里代码如下:#include <cstdio>#include <cstring>#include <vector>#include<iostream>#include <algorithm>#define maxd 2020using namespace std;int n,m,tot=0;//n个点,m条边 struct node{ int to,next;}edge[500050];int nume.
图论:KM算法求解二分图最大权完美匹配
#include<iostream>//二分图最大权匹配 #include<cstdio>#include<algorithm>#define N 505#define M 250005#define INF 9990365505#define ll long longusing namespace std;int n,m,x,y,z,tot,tim,l,r,q[N],fr[N],nxt[M],d1[M],d2[M];int pre[N],visx[N]
spfa最长路差分约束
题目链接:点击这里#include<cstring>#include<iostream>#include<queue>#include<cstdio>#include<algorithm>using namespace std;inline int read(){ int a=0;char x=getchar(); while(x<'0'||x>'9')x=getchar(); while(x>
图论:spfa求同余最短路
题目来自:点击这里#include<cstring>#include<iostream>#include<queue>#include<cstdio>#include<algorithm>using namespace std;typedef long long ll;const ll inf = 0x3f3f3f3f;struct edge{ ll to,w,next;}kkk[1000050*2];ll head[10
数论:Stirling数
第一类Stirling数有n个仓库,每个仓库有2把钥匙,共2n把钥匙,有n位保管员,保管员分别属于k个不同的部,部中的保管员数量和他们管理的仓库数量一样多,例如第i个部有m个管理员,保管m个仓库,如何放钥匙,使得同部的所有保管员能打开本部的所有仓库,但是无法打开其他的仓库。s[n][k]=s[n-1][k-1]+(n-1)s[n-1][k] ,1<=k<=ns[0][0]=1,s[k][0]=0 , 1<=k<=n第二类Stirling数把n个不同的球分配到k个相同的盒子里
博弈论:ccpc2020nim
#include<cstdio>#include<algorithm>#include<iostream>#include<vector>#include<cstring>#include<cmath>using namespace std;typedef long long ll;const int mxm=2e5+5;bool isprime[mxm];int num_pri=0;int prime[mxm]
数论:min25筛求素数个数
#include<cstdio>#include<algorithm>#include<iostream>#include<vector>#include<cstring>#include<cmath>using namespace std;typedef long long ll;const int mxm=2e5+5;const ll N=1e10+5;bool isprime[mxm];ll prime[
数论:min25筛求n个1相加
#include<cstdio>#include<algorithm>#include<iostream>#include<vector>#include<cstring>#include<cmath>using namespace std;typedef long long ll;const int mxm=2e5+5;const ll N=1e10+5;bool isprime[mxm];ll prime[
数论:min25筛求sum(1,n)
#include<cstdio>#include<algorithm>#include<iostream>#include<vector>#include<cstring>#include<cmath>using namespace std;typedef long long ll;const int mxm=2e5+5;const ll N=1e10+5;bool isprime[mxm];ll prime[
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