题目描述 In some countries building highways takes a lot of time- Maybe that's because there are many possiblities to construct a network of highways and engineers can't make up their minds which one to choose. Suppose we have a list of cities that can b…

[题目分析] Matrix-Tree定理+高斯消元 求矩阵行列式的值,就可以得到生成树的个数. 至于证明,可以去看Vflea King(炸树狂魔)的博客 [代码] #include <cmath> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define eps 1e-8 #define ma…

HIGH - Highways no tags  In some countries building highways takes a lot of time... Maybe that's because there are many possiblities to construct a network of highways and engineers can't make up their minds which one to choose. Suppose we have a lis…

In some countries building highways takes a lot of time... Maybe that's because there are many possiblities to construct a network of highways and engineers can't make up their minds which one to choose. Suppose we have a list of cities that can be c…

3143: [Hnoi2013]游走 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 2264  Solved: 987[Submit][Status][Discuss] Description 一个无向连通图,顶点从1编号到N,边从1编号到M. 小Z在该图上进行随机游走,初始时小Z在1号顶点,每一步小Z以相等的概率随机选 择当前顶点的某条边,沿着这条边走到下一个顶点,获得等于这条边的编号的分数.当小Z 到达N号顶点时游走结束,总分为所有获得的分数…

3270: 博物馆 Time Limit: 30 Sec  Memory Limit: 128 MBSubmit: 292  Solved: 158[Submit][Status][Discuss] Description 有一天Petya和他的朋友Vasya在进行他们众多旅行中的一次旅行,他们决定去参观一座城堡博物馆.这座博物馆有着特别的样式.它包含由m条走廊连接的n间房间,并且满足可以从任何一间房间到任何一间别的房间. 两个人在博物馆里逛了一会儿后两人决定分头行动,去看各自感兴趣的艺术品.他…

EXTENDED LIGHTS OUT Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9612   Accepted: 6246 Description In an extended version of the game Lights Out, is a puzzle with 5 rows of 6 buttons each (the actual puzzle has 5 rows of 5 buttons eac…

Description 有一个球形空间产生器能够在n维空间中产生一个坚硬的球体.现在,你被困在了这个n维球体中,你只知道球 面上n+1个点的坐标,你需要以最快的速度确定这个n维球体的球心坐标,以便于摧毁这个球形空间产生器. Input 第一行是一个整数n(1<=N=10).接下来的n+1行,每行有n个实数,表示球面上一点的n维坐标.每一个实数精确到小数点 后6位,且其绝对值都不超过20000. Output 有且只有一行,依次给出球心的n维坐标(n个实数),两个实数之间用一个空格隔开.每个实数精…

Description 一个黑白网格,点一次会改变这个以及与其连通的其他方格的颜色,求最少点击次数使得所有全部变成黑色. Sol 高斯消元解异或方程组. 先建立一个方程组. \(x_i\) 表示这个点是否被用过. 因为第二次使用同一个点,这个点的贡献就被消除了,所以每个点只会被用 0/1 次. \(a_{ij}\) 表示 \(j\) 点对 \(i\) 是否有影响,有影响为 1 否则为 0. 最后的一位表示最后的状态^最初的状态. 这样就列出来了 \(n*m\) 个方程组,一共 \(n*m\) 个…

[题目分析] 高斯消元求线性基. 题目本身不难,但是两种维护线性基的方法引起了我的思考. void gauss(){ k=n; F(i,1,n){ F(j,i+1,n) if (a[j]>a[i]) swap(a[i],a[j]); if (!a[i]) {k=i-1; break;} D(j,30,0) if (a[i]>>j & 1){ b[i]=j; F(x,1,n) if (x!=i && a[x]>>j&1) a[x]^=a[i];…