一、Dijkstra最短路径算法
是从一个顶点到其余各顶点的最短路径算法，解决的是有向图中最短路径问题。迪杰斯特拉算法主要特点是以起始点为中心向外层层扩展，直到扩展到终点为止。
实现一
//
// Dijkstra
// ACM
// Find the number of minimal path
//
// Created by Rachel on 18-2-12.
// Copyright (c) 2014年 ZJU. All rights reserved.
//
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <functional>
#include <utility>
#include <memory.h>
using namespace std;
#define N 505
#define INF 100000000
#define min(a,b) a<b?a:b
#define max(a,b) a>b?a:b
int map[N][N];
int minres[N]; //min distance from source to point_i
bool visited[N];
int weight[N];
void init(int n)
{
int i,j;
for (i=0; i<n; i++) {
for (j=0; j<n; j++) {
map[i][j] = INF;
}
minres[i] = INF;
}
memset(visited, false, sizeof(visited));
}
void dijkstra(int source, int dest, int n)
{
int i,j;
for(i=0;i<n;i++)
minres[i]=map[source][i];
visited[source]=true;
// (n-1) times, each time select one point into the start point set
for (j=0; j<n-1; j++) {
//select a point to add into the start point set
int minn = INF, point=-1;
for(i=0;i<n;i++)
if (!visited[i]&&minres[i]<minn) {
minn = minres[i];
point = i;
}
visited[point] = true;
//update the min distance of other points
for (i=0; i<n; i++) {
if (!visited[i]&&minres[i]>minres[point]+map[point][i]) {
minres[i] = minres[point]+map[point][i];
}
}
}
}
void dfs(int source, int dest,int n, int curpoint, int curdis, int cursum, int* num, int* sum)
{
if (curpoint==dest && minres[dest]==curdis) {
*num = *num+1;
*sum = max(*sum, cursum);
return;
}
if (curdis>minres[dest])
return;
for (int i=0; i<n; i++) {
if(!visited[i]&&map[curpoint][i]!=INF)
{
visited[i] = true;
dfs(source, dest, n, i, curdis+map[curpoint][i], cursum+weight[i], num, sum);
visited[i] = false;
}
}
}
int main()
{
int i,m,n,a,b,t,source,dest;
while (cin>>n>>m) {
cin>>source>>dest;
for (i=0; i<n; i++) {
cin>>weight[i]; //#peoples @ each point
}
init(n);
for(i=0;i<m;i++)
{
scanf("%d%d%d",&a,&b,&t);
map[b][a] = map[a][b]= min(map[a][b],t);
}
dijkstra(source,dest,n);
minres[source] = 0;
int num = 0, sum = 0;
memset(visited, false, sizeof(visited));
visited[source] = true;
dfs(source, dest, n, source, 0, weight[source], &num, &sum);
cout<<num<<" "<<sum<<endl;
}
}
实现二
/*Dijkstra求单源最短路径 2010.8.26*/
/*http://www.cnblogs.com/dolphin0520/archive/2011/08/26/2155202.html
*/
#include <iostream>
#include<stack>
#define M 100
#define N 100
using namespace std;
typedef struct node
{
int matrix[N][M]; //邻接矩阵
int n; //顶点数
int e; //边数
}MGraph;
void DijkstraPath(MGraph g,int *dist,int *path,int v0) //v0表示源顶点
{
int i,j,k;
bool *visited=(bool *)malloc(sizeof(bool)*g.n);
for(i=0;i<g.n;i++) //初始化
{
if(g.matrix[v0][i]>0&&i!=v0)
{
dist[i]=g.matrix[v0][i];
path[i]=v0; //path记录最短路径上从v0到i的前一个顶点
}
else
{
dist[i]=INT_MAX; //若i不与v0直接相邻，则权值置为无穷大
path[i]=-1;
}
visited[i]=false;
path[v0]=v0;
dist[v0]=0;
}
visited[v0]=true;
for(i=1;i<g.n;i++) //循环扩展n-1次
{
int min=INT_MAX;
int u;
for(j=0;j<g.n;j++) //寻找未被扩展的权值最小的顶点
{
if(visited[j]==false&&dist[j]<min)
{
min=dist[j];
u=j;
}
}
visited[u]=true;
for(k=0;k<g.n;k++) //更新dist数组的值和路径的值
{
if(visited[k]==false&&g.matrix[u][k]>0&&min+g.matrix[u][k]<dist[k])
{
dist[k]=min+g.matrix[u][k];
path[k]=u;
}
}
}
}
void showPath(int *path,int v,int v0) //打印最短路径上的各个顶点
{
stack<int> s;
int u=v;
while(v!=v0)
{
s.push(v);
v=path[v];
}
s.push(v);
while(!s.empty())
{
cout<<s.top()<<" ";
s.pop();
}
}
int main(int argc, char *argv[])
{
int n,e; //表示输入的顶点数和边数
while(cin>>n>>e&&e!=0)
{
int i,j;
int s,t,w; //表示存在一条边s->t,权值为w
MGraph g;
int v0;
int *dist=(int *)malloc(sizeof(int)*n);
int *path=(int *)malloc(sizeof(int)*n);
for(i=0;i<N;i++)
for(j=0;j<M;j++)
g.matrix[i][j]=0;
g.n=n;
g.e=e;
for(i=0;i<e;i++)
{
cin>>s>>t>>w;
g.matrix[s][t]=w;
}
cin>>v0; //输入源顶点
DijkstraPath(g,dist,path,v0);
for(i=0;i<n;i++)
{
if(i!=v0)
{
showPath(path,i,v0);
cout<<dist[i]<<endl;
}
}
}
return 0;
}
二、基于bitset排序
各种排序算法，生成随机文件程序。
//purpose: 生成随机的不重复的测试数据
//copyright@ 2011.04.19 yansha
//1000w数据量，要保证生成不重复的数据量，一般的程序没有做到。
//但，本程序做到了。
//July、2010.05.30。
#include <iostream>
#include <time.h>
#include <assert.h>
using namespace std;
const int size = 10000000;
int num[size];
int main()
{
int n;
FILE *fp = fopen("data.txt", "w");
assert(fp);
for (n = 1; n <= size; n++)
//之前此处写成了n=0;n<size。导致下面有一段小程序的测试数据出现了0，特此订正。
num[n] = n;
srand((unsigned)time(NULL));
int i, j;
for (n = 0; n < size; n++)
{
i = (rand() * RAND_MAX + rand()) % 10000000;
j = (rand() * RAND_MAX + rand()) % 10000000;
swap(num[i], num[j]);
}
for (n = 0; n < size; n++)
fprintf(fp, "%d ", num[n]);
fclose(fp);
return 0;
}
基于bitset实现方法
#include<iostream>
#include<bitset>
#include<assert.h>
#include<time.h>
using namespace std;
const int max_each_scan=5000000;
int main(){
clock_t begin=clock();
bitset<max_each_scan> bitmap;
bitmap.reset();
FILE* fp_unsort_file=fopen("data.txt","r");
assert(fp_unsort_file);
int num;
while(fscanf(fp_unsort_file,"%d",&num)!=EOF){
if(num<max_each_scan)
bitmap.set(num,1);
}
FILE* fp_sort_file=fopen("sort.txt","w");
assert(fp_sort_file);
int i ;
for(i=0;i<max_each_scan;i++){
if(bitmap[i]==1)
fprintf(fp_sort_file, "%d
", i);
}
int result=fseek(fp_unsort_file,0,SEEK_SET);
if(result)
cout<<"failed"
else{
bitmap.reset();
while(fscanf(fp_unsort_file,"%d",$num)!=EOF){
if(num>max_each_scan&&num<10000000){
num=num-max_ean_scan;
bitmap.set(num,1);
}
}
for(i=0;i<max_each_scan;i++){
if(bitmap[i]==1){
fprintf(fp_sort_file, "%d
", max_each_scan+i);
}
}
}
clock_t end=clock();
cout<<"排序用时："<<endl;
cout << (end - begin) / CLK_TCK << "s" << endl;
fclose(fp_sort_file);
fclose(fp_unsort_file);
return 0;
}
三、海量数据排序实例
//copyright@ yansha
//July、updated，2011.05.28。
#include <iostream>
#include <string>
#include <algorithm>
#include <time.h>
using namespace std;
int sort_num = 10000000;
int memory_size = 250000;
//每次只对250k个小数据量进行排序
int read_data(FILE *fp, int *space)
{
int index = 0;
while (index < memory_size && fscanf(fp, "%d ", &space[index]) != EOF)
index++;
return index;
}
void write_data(FILE *fp, int *space, int num)
{
int index = 0;
while (index < num)
{
fprintf(fp, "%d ", space[index]);
index++;
}
}
// check the file pointer whether valid or not.
void check_fp(FILE *fp)
{
if (fp == NULL)
{
cout << "The file pointer is invalid!" << endl;
exit(1);
}
}
int compare(const void *first_num, const void *second_num)
{
return *(int *)first_num - *(int *)second_num;
}
string new_file_name(int n)
{
char file_name[20];
sprintf(file_name, "data%d.txt", n);
return file_name;
}
int memory_sort()
{
// open the target file.
FILE *fp_in_file = fopen("data.txt", "r");
check_fp(fp_in_file);
int counter = 0;
while (true)
{
// allocate space to store data read from file.
int *space = new int[memory_size];
int num = read_data(fp_in_file, space);
// the memory sort have finished if not numbers any more.
if (num == 0)
break;
// quick sort.
qsort(space, num, sizeof(int), compare);
// create a new auxiliary file name.
string file_name = new_file_name(++counter);
FILE *fp_aux_file = fopen(file_name.c_str(), "w");
check_fp(fp_aux_file);
// write the orderly numbers into auxiliary file.
write_data(fp_aux_file, space, num);
fclose(fp_aux_file);
delete []space;
}
fclose(fp_in_file);
// return the number of auxiliary files.
return counter;
}
void merge_sort(int file_num)
{
if (file_num <= 0)
return;
// create a new file to store result.
FILE *fp_out_file = fopen("result.txt", "w");
check_fp(fp_out_file);
// allocate a array to store the file pointer.
FILE **fp_array = new FILE *[file_num];
int i;
for (i = 0; i < file_num; i++)
{
string file_name = new_file_name(i + 1);
fp_array[i] = fopen(file_name.c_str(), "r");
check_fp(fp_array[i]);
}
int *first_data = new int[file_num];
//new出个大小为0.1亿/250k数组，由指针first_data指示数组首地址
bool *finish = new bool[file_num];
memset(finish, false, sizeof(bool) * file_num);
// read the first number of every auxiliary file.
for (i = 0; i < file_num; i++)
fscanf(fp_array[i], "%d ", &first_data[i]);
while (true)
{
int index = 0;
while (index < file_num && finish[index])
index++;
// the finish condition of the merge sort.
if (index >= file_num)
break;
//主要的修改在上面两行代码，就是merge sort结束条件。
//要保证所有文件都读完，必须使得finish[0]...finish[40]都为真
//July、yansha，555，2011.05.29。
int min_data = first_data[index];
// choose the relative minimum in the array of first_data.
for (i = index + 1; i < file_num; i++)
{
if (min_data > first_data[i] && !finish[i])
//一旦发现比min_data更小的数据first_data[i]
{
min_data = first_data[i];
//则置min_data<-first_data[i]index = i;
//把下标i 赋给index。
}
}
// write the orderly result to file.
fprintf(fp_out_file, "%d ", min_data);
if (fscanf(fp_array[index], "%d ", &first_data[index]) == EOF)
finish[index] = true;
}
fclose(fp_out_file);
delete []finish;
delete []first_data;
for (i = 0; i < file_num; i++)
fclose(fp_array[i]);
delete [] fp_array;
}
int main()
{
clock_t start_memory_sort = clock();
int aux_file_num = memory_sort();
clock_t end_memory_sort = clock();
cout << "The time needs in memory sort: " << end_memory_sort - start_memory_sort << endl;
clock_t start_merge_sort = clock();
merge_sort(aux_file_num);
clock_t end_merge_sort = clock();
cout << "The time needs in merge sort: " << end_merge_sort - start_merge_sort << endl;
system("pause");
return 0;
}
四、多路归并排序
//copyright@ 纯净的天空 && yansha
//5、July，updated，2010.05.28。
//harryshayne，update again。2011.6.30
#include <iostream>
#include <ctime>
#include <fstream>
//#include "ExternSort.h"
using namespace std;
//使用多路归并进行外排序的类
//ExternSort.h
/*
* 大数据量的排序
* 多路归并排序
* 以千万级整数从小到大排序为例
* 一个比较简单的例子，没有建立内存缓冲区
*/
#ifndef EXTERN_SORT_H
#define EXTERN_SORT_H
#include <cassert>
//#define k 5
#define MIN -1//这里开始的时候出现了一个BUG，如果定义的MIN大于等于待排序的数，则会是算法出现错误
#define MAX 10000000//最大值，附加在归并文件结尾
typedef int* LoserTree;
typedef int* External;
class ExternSort
{
public:
void sort()
{
time_t start = time(NULL);
//将文件内容分块在内存中排序，并分别写入临时文件
k = memory_sort(); //
//归并临时文件内容到输出文件
//merge_sort(file_count);
ls=new int[k];
b=new int[k+1];
K_Merge();
delete []ls;
delete []b;
time_t end = time(NULL);
printf("total time:%f
", (end - start) * 1000.0/ CLOCKS_PER_SEC);
}
//input_file:输入文件名
//out_file:输出文件名
//count: 每次在内存中排序的整数个数
ExternSort(const char *input_file, const char * out_file, int count)
{
m_count = count;
m_in_file = new char[strlen(input_file) + 1];
strcpy(m_in_file, input_file);
m_out_file = new char[strlen(out_file) + 1];
strcpy(m_out_file, out_file);
}
virtual ~ExternSort()
{
delete [] m_in_file;
delete [] m_out_file;
}
private:
int m_count; //数组长度
char *m_in_file; //输入文件的路径
char *m_out_file; //输出文件的路径
int k;//归并数，此数必须要内排序之后才能得到，所以下面的ls和b都只能定义为指针(注意和书上区别)
LoserTree ls;//定义成为指针，之后动态生成数组
External b;//定义成为指针，在成员函数中可以把它当成数组使用
//int External[k];
protected:
int read_data(FILE* f, int a[], int n)
{
int i = 0;
while(i < n && (fscanf(f, "%d", &a[i]) != EOF)) i++;
printf("read:%d integer
", i);
return i;
}
void write_data(FILE* f, int a[], int n)
{
for(int i = 0; i < n; ++i)
fprintf(f, "%d ", a[i]);
fprintf(f,"%d",MAX);//在最后写上一个最大值
}
char* temp_filename(int index)
{
char *tempfile = new char[100];
sprintf(tempfile, "temp%d.txt", index);
return tempfile;
}
static int cmp_int(const void *a, const void *b)
{
return *(int*)a - *(int*)b;
}
int memory_sort()
{
FILE* fin = fopen(m_in_file, "rt");
int n = 0, file_count = 0;
int *array = new int[m_count];
//每读入m_count个整数就在内存中做一次排序，并写入临时文件
while(( n = read_data(fin, array, m_count)) > 0)
{
qsort(array, n, sizeof(int), cmp_int);
//这里，调用了库函数阿，在第四节的c实现里，不再调用qsort。
char *fileName = temp_filename(file_count++);
FILE *tempFile = fopen(fileName, "w");
free(fileName);
write_data(tempFile, array, n);
fclose(tempFile);
}
delete [] array;
fclose(fin);
return file_count;
}
void Adjust(int s)
{//沿从叶子节点b[s]到根节点ls[0]的路径调整败者树
int t=(s+k)/2;//ls[t]是b[s]的双亲节点
while(t>0)
{
if(b[s]>b[ls[t]])//如果失败，则失败者位置s留下，s指向新的胜利者
{
int tmp=s;
s=ls[t];
ls[t]=tmp;
}
t=t/2;
}
ls[0]=s;//ls[0]存放调整后的最大值的位置
}
void CreateLoserTree()
{
b[k]=MIN;//额外的存储一个最小值
for(int i=0;i<k;i++)ls[i]=k;//先初始化为指向最小值，这样后面的调整才是正确的
//这样能保证非叶子节点都是子树中的“二把手”
for(i=k-1;i>=0;i--)
Adjust(i);//依次从b[k-1],b[k-2]...b[0]出发调整败者树
}
void K_Merge()
{//利用败者数把k个输入归并段归并到输出段中
//b中前k个变量存放k个输入段中当前记录的元素
//归并临时文件
FILE *fout = fopen(m_out_file, "wt");
FILE* *farray = new FILE*[k];
int i;
for(i = 0; i < k; ++i) //打开所有k路输入文件
{
char* fileName = temp_filename(i);
farray[i] = fopen(fileName, "rt");
free(fileName);
}
for(i = 0; i < k; ++i) //初始读取
{
if(fscanf(farray[i], "%d", &b[i]) == EOF)//读每个文件的第一个数到data数组
{
printf("there is no %d file to merge!"，k);
return;
}
}
// for(int i=0;i<k;i++)input(b[i]);
CreateLoserTree();
int q;
while(b[ls[0]]!=MAX)//
{
q=ls[0];//q用来存储b中最小值的位置，同时也对应一路文件
//output(q);
fprintf(fout,"%d ",b[q]);
//input(b[q],q);
fscanf(farray[q],"%d",&b[q]);
Adjust(q);
}
//output(ls[0]);
fprintf(fout,"%d ",b[ls[0]]);
//delete [] hasNext;
//delete [] data;
for(i = 0; i < k; ++i) //清理工作
{
fclose(farray[i]);
}
delete [] farray;
fclose(fout);
}
/*
void merge_sort(int file_count)
{
if(file_count <= 0) return;
//归并临时文件
FILE *fout = fopen(m_out_file, "wt");
FILE* *farray = new FILE*[file_count];
int i;
for(i = 0; i < file_count; ++i)
{
char* fileName = temp_filename(i);
farray[i] = fopen(fileName, "rt");
free(fileName);
}
int *data = new int[file_count];//存储每个文件当前的一个数字
bool *hasNext = new bool[file_count];//标记文件是否读完
memset(data, 0, sizeof(int) * file_count);
memset(hasNext, 1, sizeof(bool) * file_count);
for(i = 0; i < file_count; ++i) //初始读取
{
if(fscanf(farray[i], "%d", &data[i]) == EOF)//读每个文件的第一个数到data数组
hasNext[i] = false;
}
while(true) //循环读取和输出，选择最小数的方法是简单遍历选择法
{
//求data中可用的最小的数字，并记录对应文件的索引
int min = data[0];
int j = 0;
while (j < file_count && !hasNext[j]) //顺序跳过已读取完毕的文件
j++;
if (j >= file_count) //没有可取的数字，终止归并
break;
for(i = j +1; i < file_count; ++i) //选择最小数，这里应该是i=j吧！但结果是一样的！
{
if(hasNext[i] && min > data[i])
{
min = data[i];
j = i;
}
}
if(fscanf(farray[j], "%d", &data[j]) == EOF) //读取文件的下一个元素
hasNext[j] = false;
fprintf(fout, "%d ", min);
}
delete [] hasNext;
delete [] data;
for(i = 0; i < file_count; ++i)
{
fclose(farray[i]);
}
delete [] farray;
fclose(fout);
}
*/
};
#endif
//测试主函数文件
/*
* 大文件排序
* 数据不能一次性全部装入内存
* 排序文件里有多个整数，整数之间用空格隔开
*/
const unsigned int count = 10000000; // 文件里数据的行数
const unsigned int number_to_sort = 100000; //在内存中一次排序的数量
const char *unsort_file = "unsort_data.txt"; //原始未排序的文件名
const char *sort_file = "sort_data.txt"; //已排序的文件名
void init_data(unsigned int num); //随机生成数据文件
int main(int argc, char* *argv)
{
srand(time(NULL));
init_data(count);
ExternSort extSort(unsort_file, sort_file, number_to_sort);
extSort.sort();
system("pause");
return 0;
}
void init_data(unsigned int num)
{
FILE* f = fopen(unsort_file, "wt");
for(int i = 0; i < num; ++i)
fprintf(f, "%d ", rand());
fclose(f);
}
五、字符串回文结构判断
class Solution{
//http://blog.csdn.net/v_july_v/article/details/6712171
public:
/**
*check weather s is a palindrome, n is the length of string s
*Copyright(C) fairywell 2011
*/
bool IsPalindrome(const char *s, int n)
{
if (s == 0 || n < 1) return false; // invalid string
char *front, *back;
front = s; back = s + n - 1; // set front and back to the begin and endof the string
while (front < back) {
if (*front != *back) return false; // not a palindrome
++front; --back;
}
return true; // check over, it's a palindrome
}
/**
*check weather s is a palindrome, n is the length of string s
*Copyright(C) fairywell 2011
*/
bool IsPalindrome2(const char *s, int n)
{
if (s == 0 || n < 1) return false; // invalid string
char *first, *second;
int m = ((n>>1) - 1) >= 0 ? (n>>1) - 1 : 0; // m is themiddle point of s
first = s + m; second = s + n - 1 - m;
while (first >= s)
if (s[first--] !=s[second++]) return false; // not equal, so it's not apalindrome
return true; // check over, it's a palindrome
}
/**
*find the longest palindrome in a string, n is the length of string s
*Copyright(C) fairywell 2011
*/
int LongestPalindrome(const char *s, int n)
{
int i, j, max;
if (s == 0 || n < 1) return 0;
max = 0;
for (i = 0; i < n; ++i) { // i is the middle point of the palindrome
for (j = 0; (i-j >= 0) && (i+j < n); ++j) // if the lengthof the palindrome is odd
if (s[i-j] != s[i+j]) break;
if (j*2+1 > max) max = j * 2 + 1;
for (j = 0; (i-j >= 0) && (i+j+1 < n); ++j) // for theeven case
if (s[i-j] != s[i+j+1]) break;
if (j*2+2 > max) max = j * 2 + 2;
}
return max;
}
int LongestPalindrome(const char *s, int n)
{
int max = 0;
int i,j;
for ( i = 0; i < n ; i++ )
{//以i为中心开始计算回文子串
//计算奇数回问子串长度
for ( j = 0; (i-j) >= 0 && (i+j) < n; j++ )
{
if ( s[i-j] != s[i+j] )
{
break;
}
else
{
max = GETMAX(max, (2 * j + 1));
}
}
//计算偶数回问子串长度
for ( j = 0; (i-j) >= 0 && i + j + 1< n; j++ )
{
if ( s[i-j] != s[i+j+1])
{
break;
}
else
{
max = GETMAX(max, ( 2 * j + 2) );
}
}
}
return max;
}
}