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www.eeworm.com/read/409150/11344420
m treematrix.m
clear;
n=30;
x=rand(n,2);
clf;
for i=1:n
for j=1:n
w(i,j)=(x(i,1)-x(j,1))^2+(x(i,2)-x(j,2))^2;
end
w(i,i)=100;
end
hold on;axis([0,1,0,1]);
for i=1:n
plot(x(i,1),
www.eeworm.com/read/408902/11365624
py testhadoop.py
#Licensed to the Apache Software Foundation (ASF) under one
#or more contributor license agreements. See the NOTICE file
#distributed with this work for additional information
#regarding copyright ow
www.eeworm.com/read/407780/11410823
txt 求多短路的最短路径，用动态规划法.txt
#include
//求多短路的最短路径，用动态规划法
/*
输入
10 18
0 1 4
0 2 2
0 3 3
1 4 9
1 5 8
2 4 6
2 5 7
2 6 8
3 5 4
3 6 7
4 7 5
4 8 6
5 7 8
5 8 6
6 7 6
6 8 5
7 9 7
8 9 3
输出
0->3->5-
www.eeworm.com/read/407780/11410865
txt 判断有向图是否有环.txt
#include
#include
#include
#include
using namespace std;
#define NMAX 105
int path[NMAX][NMAX];
int Ingree[NMAX];//各个点的入度
//判断一个有向图是否有环
/*
输入：
4
www.eeworm.com/read/407601/11414334
dpr ymlite.dpr
program YmLite;
uses
Windows,
Forms,
Dialogs,
classes,
Sysutils,
FileCtrl,
StdCtrls,
YchatU1 in 'YchatU1.pas' {Form1},
YchatU2 in 'YchatU2.pas' {Form2},
YchatU3 in '
www.eeworm.com/read/405817/11456276
cpp the unique mst(prime，次小生成树判定).cpp
#include
#include
#include
#include
#include
using namespace std;
const int MAX = 110;
int t,n,m;
int path[MAX][MAX];
int dmax[MAX][MAX];
str
www.eeworm.com/read/405817/11456313
cpp network(割点).cpp
#include
#include
#include
#include
#include
using namespace std;
const int MAX = 110;
int n;
bool path[MAX][MAX];
int vis[MAX], low[MAX];
www.eeworm.com/read/405817/11456338
cpp xyzzy(bellman ford).cpp
//中间不能死
//有正权环的话，得找出终点是否受环影响
#include
#include
#include
#include
#include
using namespace std;
#define MAX 110
int n,m;
int v[MAX],dist[MAX],
www.eeworm.com/read/405817/11456344
cpp gopher ii(二分匹配).cpp
#include
#include
#include
#include
using namespace std;
#define NMAX 110
int n,m,s,v;
struct node {
double x,y;
}go[NMAX], hole[NMAX];
bool path[NMA
www.eeworm.com/read/405817/11456352
cpp maximum clique(最大团).cpp
#include
#include
#define NMAX 51
bool path[NMAX][NMAX];
int n, mmax;
int dp[NMAX];
bool v[NMAX];
bool dfs(int pos, int size)
{
int i, j, unvis;
bool tv[NMAX];