<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[crate_M的专栏]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/crate_M</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; crate_M]]></copyright><item><title><![CDATA[07-图6 旅游规划   (25分)]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51276365</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51276365</guid><author>crate_M</author><pubDate>Thu, 28 Apr 2016 17:37:57 +0800</pubDate><description><![CDATA[单源最短路算法#include<bits/stdc++.h>
#define inf 260000
using namespace std;
struct node{
    int dist;
    int cost;
};node Map[505][505];
int  dist[505],cost[505],n,m,s,d;int main()
{
    scanf("%d %d %d %]]></description><category></category></item><item><title><![CDATA[05-树7 堆中的路径   (25分)]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51274964</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51274964</guid><author>crate_M</author><pubDate>Thu, 28 Apr 2016 15:50:17 +0800</pubDate><description><![CDATA[ac代码#include<stdio.h>#define MaxN 1001
#define MinH -10001int H[MaxN],size;void Create()
{
    size = 0;
    H[0] = MinH;
}void Insert(int X)
{
    int i;
    for(i=++size;H[i/2]>X;i/=2)
        H[i] =]]></description><category></category></item><item><title><![CDATA[05-树8 File Transfer   (25分)]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51274953</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51274953</guid><author>crate_M</author><pubDate>Thu, 28 Apr 2016 15:48:48 +0800</pubDate><description><![CDATA[并查集#include<bits/stdc++.h>
using namespace std;
const int maxn = 1000;
int father[maxn];int findroot(int x){
    if(x != father[x]){
        father[x] = findroot(father[x]);
    }
    return father[x];]]></description><category></category></item><item><title><![CDATA[04-树6 Complete Binary Search Tree]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51274936</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51274936</guid><author>crate_M</author><pubDate>Thu, 28 Apr 2016 15:47:17 +0800</pubDate><description><![CDATA[用数组ac代码#include<cstdio>
#include<cstdlib>const int maxx = 1005;
int node[maxx];
int tree[maxx];
int pos = 0,n;int cmp(const void *a,const void *b){
    int *pa = (int *)a;
    int *pb = (int *)b;
    r]]></description><category></category></item><item><title><![CDATA[03-树3 Tree Traversals Again]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51251752</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51251752</guid><author>crate_M</author><pubDate>Tue, 26 Apr 2016 16:39:02 +0800</pubDate><description><![CDATA[根据题目: 
入栈的顺序即为先序遍历的顺序 
出栈的顺序即为中序遍历的顺序 
然后根据先序遍历、中序遍历来构建二叉树AC代码#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string>
#include<stack>using namespace std;typedef struct TreeNode *BinTre]]></description><category></category></item><item><title><![CDATA[06-图3 六度空间   (30分)]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51248094</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51248094</guid><author>crate_M</author><pubDate>Tue, 26 Apr 2016 09:54:56 +0800</pubDate><description><![CDATA[BFS 
key：记录BFS的层数，在BFS层数为6的时候计算count/Nv 
ac代码#include<stdio.h>
#include<iostream>
#include<queue>using namespace std;int G[10005][10005];
bool Visited[10005];
int Nv,Ne;void CreateGraph()
{
    scanf("]]></description><category></category></item><item><title><![CDATA[06-图2 Saving James Bond - Easy Version]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51246548</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51246548</guid><author>crate_M</author><pubDate>Mon, 25 Apr 2016 23:00:29 +0800</pubDate><description><![CDATA[#include<iostream>
#include<vector>
#include<cmath>
using namespace std;#define MinLen 42.5
struct Point{
    int x;
    int y;
    bool visited;
};
Point O;int N,D;
vector<Point> s;double Distance(Poi]]></description><category></category></item><item><title><![CDATA[06-图1 列出连通集   (25分)]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51246531</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51246531</guid><author>crate_M</author><pubDate>Mon, 25 Apr 2016 22:58:51 +0800</pubDate><description><![CDATA[BFS中非STL实现#include<cstdio>#define N 15int G[N][N],Nv,Ne;
bool Visited[N];void InitVisit()
{
    for(int i=0;i<N;i++)
        Visited[i] = false;
}void DFS(int V)
{
    Visited[V] = true;
    printf("%d]]></description><category></category></item><item><title><![CDATA[PAT 1074 Reversing Linked List]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51232566</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51232566</guid><author>crate_M</author><pubDate>Sun, 24 Apr 2016 11:08:07 +0800</pubDate><description><![CDATA[题目链接AC代码#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;int list[100010];
int node[100010][2];int main()
{
//  freopen("F:\\in1.txt","r",stdin);
    int str,num,r;
    scan]]></description><category></category></item><item><title><![CDATA[蓝桥杯~快速排列]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51160196</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51160196</guid><author>crate_M</author><pubDate>Fri, 15 Apr 2016 11:20:54 +0800</pubDate><description><![CDATA[快速排序排序在各种场合经常被用到。 
快速排序是十分常用的高效率的算法。其思想是：先选一个“标尺”， 
用它把整个队列过一遍筛子， 
以保证：其左边的元素都不大于它，其右边的元素都不小于它。这样，排序问题就被分割为两个子区间。 
再分别对子区间排序就可以了。下面的代码是一种实现，请分析并填写划线部分缺少的代码。#include<stdio.h>void swap(int a[],int i,int]]></description><category></category></item><item><title><![CDATA[蓝桥杯~凑算式]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51159936</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51159936</guid><author>crate_M</author><pubDate>Fri, 15 Apr 2016 11:00:26 +0800</pubDate><description><![CDATA[全排列生成1~9的数字然后判断是否满足条件#include<stdio.h>
int ans = 0;
int num[10];
int visit[10];
void Solve()  
{  
    double sum = num[0] + (double)num[1] / num[2] + (double)(num[3]*100+num[4]*10+num[5])/(num[6]*100+]]></description><category></category></item><item><title><![CDATA[03-树2 List Leaves   (25分)]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51147336</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51147336</guid><author>crate_M</author><pubDate>Wed, 13 Apr 2016 22:13:38 +0800</pubDate><description><![CDATA[在网上看到别人的代码，真心觉得写的很好#include<stdio.h>
#include<ctype.h>struct Node{
    int root;
    int left;
    int right;
};int main(){
    //create the tree
    int n;
    struct Node nodes[10];
    scanf("%d",&n]]></description><category></category></item><item><title><![CDATA[pat 1009]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51130267</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51130267</guid><author>crate_M</author><pubDate>Tue, 12 Apr 2016 09:12:36 +0800</pubDate><description><![CDATA[多项式乘法: 
用一个数组储存相乘后的到的元素 
使用double型的数组，下标表示指数，対应的值表示系数。#include<stdio.h>
#include<math.h>typedef struct{
    int expon;
    double coef;
}poly;poly a[15],b[15];
double ans[2008];
int k1,k2,cnt=0;void In]]></description><category></category></item><item><title><![CDATA[Pat 1002]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51126255</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51126255</guid><author>crate_M</author><pubDate>Mon, 11 Apr 2016 22:10:13 +0800</pubDate><description><![CDATA[多项式加法~数组实现 
直接贴代码了#include<stdio.h>typedef struct ploynode{
    int expon;
    double coef;
}poly;int main()
{
    int k1,k2;
    poly a[12],b[12],c[30];
    int i,j,pos=0;    scanf("%d",&k1);
    for(]]></description><category></category></item><item><title><![CDATA[计算机体系结构与参考模型]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51010883</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51010883</guid><author>crate_M</author><pubDate>Wed, 30 Mar 2016 08:43:02 +0800</pubDate><description><![CDATA[计算机网络分层结构 
为何要采用分层结构: 
(1)发起通讯的计算机必须将数据通信的通道激活,所谓激活就是发送一些指令。 
(2)要告诉网络如何识别接受的计算机 
(3)发起通讯的计算机必须查明对方计算机是否已经开机，并且与网络连接正常 
(4)发起通讯的计算机的计算机中的应用程序必须弄清楚对方的文件管理程序是否已经做好接受文件和储存文件的准备工作 
(5)如果计算机文件格式不兼容,则至少其中的一台]]></description><category></category></item><item><title><![CDATA[计算机网络概述]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/51003631</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/51003631</guid><author>crate_M</author><pubDate>Tue, 29 Mar 2016 10:24:23 +0800</pubDate><description><![CDATA[一.计算机网络的概念、组成、功能计算机网络:一些相互连接的、自治的计算机的集合。计算机网络组成: 
(1)边缘部分:所有连接在Internet上的主机组成，用户直接使用，用来进行通讯和资源共享(2)核心部分:由大量网络和连接这些网络的路由器组成，这部分是为边缘部分提供服务的计算机网络的功能:(1)连通性(2)资源共享二.计算机网络的分类1.按网络的作用范围分类: 
(1)广域网(WAN) Wide]]></description><category></category></item><item><title><![CDATA[回溯法~相关内容]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/50996756</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/50996756</guid><author>crate_M</author><pubDate>Mon, 28 Mar 2016 11:49:16 +0800</pubDate><description><![CDATA[回溯法:在递归的构建中，生成和检查过程就可以有机的结合起来，从而减少不必要的枚举排列枚举的两种方式void dfs(int index)
{
    if(index = 9)
    {
        Solve();
        return;
    }    for(int i=1;i<10;i++)
    {
        if(!visit[i])
        {]]></description><category></category></item><item><title><![CDATA[C ++排序函数sort(),qsort()]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/50336817</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/50336817</guid><author>crate_M</author><pubDate>Thu, 17 Dec 2015 04:11:55 +0800</pubDate><description><![CDATA[sort():给指定区间的所有元素进行排序example这里写代码片#include<stdio.h>
#include<algorithm>using namespace std;int main()
{
    int a[5] = {1,5,4,2,3};
    sort(a,a+5);
    printf("%d",a[0]);
    for(int i=1;i<5;i++)]]></description><category></category></item><item><title><![CDATA[PAT 1045 快速排序]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/50255969</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/50255969</guid><author>crate_M</author><pubDate>Thu, 10 Dec 2015 22:33:06 +0800</pubDate><description><![CDATA[快速排序 
题目链接先用快速排序对数组进行排序，对原数组进行遍历 
如果满足该数满足他是遍历过程中最大的，而且该数和排好序的数组中对应的元素相等就说名他是主元int comp(const void *a,const void *b)
{
    return *（int *)a - *(int *)b;
}
qsort(arr_s,N,sizeof(int),comp)]]></description><category></category></item><item><title><![CDATA[1049 数据片段和]]></title><link>https://blog.youkuaiyun.com/crate_M/article/details/50219287</link><guid>https://blog.youkuaiyun.com/crate_M/article/details/50219287</guid><author>crate_M</author><pubDate>Tue, 08 Dec 2015 15:10:01 +0800</pubDate><description><![CDATA[给定一个正数数列，我们可以从中截取任意的连续的几个数，称为片段。例如，给定数列{0.1, 0.2, 0.3, 0.4}，我们有(0.1) (0.1, 0.2) (0.1, 0.2, 0.3) (0.1, 0.2, 0.3, 0.4) (0.2) (0.2, 0.3) (0.2, 0.3, 0.4) (0.3) (0.3, 0.4) (0.4) 这10个片段。给定正整数数列，求出全部片段包含的所有的数]]></description><category></category></item></channel></rss>