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經典的算法問題：
（拿出來供自己和大家都看看吧，算法學習不能停。）
一． 遞歸與分冶算法。
題目：棋盤履蓋 (javascript)
<HTML>
<HEAD>
<TITLE> 棋盤履蓋 </TITLE>
</HEAD>
<Script Language="JavaScript">
function getRandColor(){
return '#'+(Math.random()*0xffffff<<0).toString(16);
}
document.write("<BODY bgcolor=" + getRandColor() +">")
document.write("<h1><center><font color =red> 棋盤履蓋 </font></center></h1>");
var tile=0;
var t = 0;
//new 個數組
</Script>
<Script Language="JavaScript">
var k = prompt(" 輸入棋盤的規格 /n2<sup>k</sup>","2");
var row = Math.pow(2,k);
var col = row;
alert(" 行數 " + row +" 列數 " + col);
var board = new Array(row);
for(i = 0; i<row; i++) {
board[i] = new Array(col);
for(j = 0; j<col; j++) {
board[i][j] = 0;
}
}
var drow = prompt(" 輸入特殊方格所在的行 ","0");
var dcol = prompt(" 輸入特殊方格所在的列 ","0");
alert("chessBoard begin");
var tile = 0;
function chessBoard(tr,tc,dr,dc,size) {
if(size == 1) {
}
else {
var t = getRandColor() + "," + (++tile);
//var t=++tile;
var s= size/2;
// 覆蓋左上角子棋盤
if(dr<tr+s && dc<tc+s) {
chessBoard(tr,tc,dr,dc,s);
}
else {
board[tr+s-1][tc+s-1] = t;
chessBoard(tr,tc,tr+s-1,tc+s-1,s);
}
// 覆蓋右上角棋盤
if(dr<tr+s && dc>=tc+s) {
chessBoard(tr,tc+s,dr,dc,s);
}
else {
board[tr+s-1][tc+s] = t;
chessBoard(tr,tc+s,tr+s-1,tc+s,s);
}
// 覆蓋左下角棋盤
if(dr>=tr+s && dc<tc+s) {
chessBoard(tr+s,tc,dr,dc,s);
}
else {
board[tr+s][tc+s-1] = t;
chessBoard(tr+s,tc,tr+s,tc+s-1,s);
}
// 覆蓋右下角棋盤
if(dr>=tr+s && dc>=tc+s) {
chessBoard(tr+s,tc+s,dr,dc,s);
}
else {
board[tr+s][tc+s] = t;
chessBoard(tr+s,tc+s,tr+s,tc+s,s);
}
}
}
chessBoard(0,0,drow,dcol,col);
board[drow][dcol] ="#ffffff" +",0";
</Script>
<table align = "center" border="0">
<Script Language = "JavaScript">
for(i = 0; i<row;i++) {
document.write("<tr>");
for(j = 0 ;j<col;j++) {
var paraArr = board[i][j].split(",");
document.write("<td width=30 height=20 bgcolor=" + paraArr[0] +">" + paraArr[1] +" </td>");
}
document.write("</tr>");
}
</Script>
</table>
</BODY>
</HTML>
格雷碼：（ java ）
import java.util.Scanner;
class Gray {
private static byte[][] ngray;
private static int num; // 輸出總行數
private int ng;
private static int loops;
public Gray(int n) {
this.ng = n;
num = (int)Math.pow(2,n);
ngray = new byte[num+1][n+1];
loops = 1;
findGrays(ng);
}
public static void findGrays(int n) {
if(n<1) return;
handleArray(n,(int)Math.pow(2,n));
findGrays(n-1);
}
private void printGrays() {
for(int i = 1; i<=num; i++) {
for(int j = 1; j<=ng;j++) {
System.out.print(ngray[i][j] + "/t");
}
System.out.println();
}
}
private static void handleArray(int n,int lo) {
int i;
// 定義一個用於處理， 0 或 1 次序的變量。
boolean zoFlag = true;
int loopnum = num/lo; // 定義一個用於控制可以循環的次數的變量
for(i = 1;i<=loopnum;++i) {
for(int j =1;j<=lo/2;j++) {
ngray[(i-1)*lo+j][n] = (byte)(zoFlag?0:1);//01 標誌爲 true ，返回 0 ，否則 1
}
for(int j = lo/2+1;j<=lo;j++) {
ngray[(i-1)*lo+j][n] = (byte)(zoFlag?1:0);//01 標誌爲 true ，返回 1 ，否則 0
}
zoFlag = zoFlag?false:true;
}
}
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
System.out.println(" 輸入格雷碼的位數 ");
int graynum = sc.nextInt();
Gray gray = new Gray(graynum);
gray.printGrays();
}
}
二． 回溯算法。
簡單 0-1 揹包問題 （ java ）
import java.util.Scanner;
public class BeiBao {
double[] price ;
double[] weight;
int n ;
double c ;
double bestprice = 0;
double currentweight = 0;
double currentprice = 0;
public void input() {
Scanner sc = new Scanner(System.in);
String inf = sc.nextLine();
String[] arr = inf.split(" ");
n = Integer.parseInt(arr[0]);
c = Double.parseDouble(arr[1]);
while (n != 0 && c != 0) {
weight = new double[n];
price = new double[n];
// 得到 weight
inf = sc.nextLine();
arr = inf.split(" ");
for (int loop = 0; loop < n; loop++) {
weight[loop] = Double.parseDouble(arr[loop]);
}
// 得到 price
inf = sc.nextLine();
arr = inf.split(" ");
for (int loop = 0; loop < n; loop++) {
price[loop] = Double.parseDouble(arr[loop]);
}
inf = sc.nextLine();
arr = inf.split(" ");
//print();
backtrack(0);
System.out.println((int)bestprice);
currentweight = currentprice = 0.0;
bestprice = 0.0;
n = Integer.parseInt(arr[0]);
c = Double.parseDouble(arr[1]);
}
sc.close();
}
public void backtrack(int i) {
if(i>=n) {
if(currentprice > bestprice) {
bestprice = currentprice;
}
return;
}
if(currentweight + weight[i] <=c ) {
currentweight += weight[i];
currentprice += price[i];
backtrack(i+1);
currentweight -= weight[i];
currentprice -= price[i];
}
if(currentprice + bound(i+1) > bestprice) {
backtrack(i+1);
}
}
double bound(int i) {
double sum = 0.0;
for(int j = i; j<n; j++) {
sum += price[j];
}
return sum;
}
void print() {
System.out.println("n" + n + "c" + c);
for(int i = 0;i<weight.length;i++) {
System.out.println(weight[i] + " " + price[i]);
}
}
public static void main(String args[]) {
BeiBao bb = new BeiBao();
bb.input();
//System.out.println(bb.bestprice);
}
}
八皇后問題 （ java ）
class EtQueen {
int n ;
int x[];
int qnum = 1;
public static void main(String args[]) {
new EtQueen(8);
}
EtQueen (int n) {
this.n = n;
x = new int[n+1];
backtrack(1);
}
void backtrack(int i) {
if(i>n) {
printQueens();
qnum ++;
}
for(int j = 1; j<=n; ++j) {
if(cangoon(i,j)) { // 符合繼續進行的條件
x[i] = j;
backtrack(i+1);
x[i] = 0;
}
}
}
boolean cangoon(int i,int j) {
for(int k = 1; k<i; k++ ) {
if(x[k] == j || Math.abs(i-k) == Math.abs(x[k]-j) ) {
return false;
}
}
return true;
}
void printQueens () {
System.out.println("No " + qnum + ":");
for(int i = 1; i<=n; i++) {
for(int j = 1; j<=n; j++) {
if(x[i] == j) {
System.out.print('A');
}else {
System.out.print('.');
}
}
System.out.println();
}
}
}
最優裝載問題 (java)
import java.util.Scanner;
class Zyzz {
int c1 ,c2 ,n ;
int[] weight;;
int bestweight ;
int currentweight ;
int r ,totalweight;
public static void main(String args[]) {
new Zyzz().input();
}
public void input() {
Scanner sc = new Scanner(System.in);
String inf;
String[] temp;
while(true) {
currentweight = bestweight = r =0;
inf = sc.nextLine();
temp = inf.split(" ");
int a,b;
a = Integer.parseInt(temp[0]);
b = Integer.parseInt(temp[1]);
n = Integer.parseInt(temp[2]);
if(a == 0 && b ==0 && n == 0) break;
if(a<b) {
c1 = a;
c2 = b;
}else {
c2 = a;
c1 = b;
}
weight = new int[n];
inf = sc.nextLine();
temp = inf.split(" ");
for(int i = 0; i<n; i++) {
weight[i] = Integer.parseInt(temp[i]);
r+=weight[i];
totalweight = r;
}
bestLoad();
if(totalweight - bestweight > c2) {
System.out.println("No");
}else {
System.out.println("Yes");
}
}
sc.close();
}
void bestLoad() {
backtrack(0);
//System.out.println("bestload = " + bestweight + "r = " +r);
}
public void backtrack(int i) {
if(i>=n) {
bestweight = currentweight ;
//System.out.println("bestweight = " + bestweight + "r = " +r);
return;
}
if(currentweight + weight[i] <=c1 ) {
r-=weight[i];
currentweight += weight[i];
//System.out.println("i " + i +"currentweight = " + currentweight + "r = " +r );
backtrack(i+1);
currentweight -= weight[i];
r+=weight[i];
}
r-=weight[i] ;
if(currentweight + r> bestweight) {
backtrack(i+1);
}
r+=weight[i];
}
}
三． 分支限界算法。
迷宮問題 (java)
import java.util.Scanner;
public class MiGong {
static char[][] maze;
static boolean[][] isVisited;
static int posStartX ;
static int posStartY ;
static int posEndX ;
static int posEndY ;
static int n = 12;// 迷宮的規模
// 先進先出對列
static Queue queue = new Queue();
static Element ende ;
static Element starte;
MiGong() {
maze = new char[n][n];
isVisited = new boolean[n][n];
Scanner sc = new Scanner(System.in);
// 初始化老鼠的位置和迷宮出口
String inf = "";
inf = sc.nextLine();
String[] posStr = inf.split(" ");
posStartX = Integer.parseInt(posStr[0])-1;
posStartY = Integer.parseInt(posStr[1])-1;
posEndX = Integer.parseInt(posStr[2])-1;
posEndY = Integer.parseInt(posStr[3])-1;
ende = new Element(posEndX,posEndY,0,null);
starte = new Element(posStartX,posStartY,0,null);
for(int i=0;i<n;i++) {
inf = sc.nextLine().trim();
for(int j = 0;j<n;j++) {
maze[i][j] = inf.charAt(j);
}
}
sc.close();
System.out.println(" 打印迷宮 ");
for(int i = 0;i<n;i++) {
for(int j = 0;j<n;j++) {
System.out.print(maze[i][j]);
isVisited[i][j] = false;
}
System.out.println();
}
}
private static class Element {
Element(int x,int y,int num,Element pre) {
this.x = x;
this.y = y;
this.num = num;
this.pre = pre;
}
int x;
int y;
Element pre;
int num;// 步數
boolean equals(Element e) {
if(e==null) return false;
if(x == e.x && y == e.y) {
return true;
}
return false;
}
void print() {
System.out.println(" 位置 " + "X=" + x + "Y=" + y + " Num=" + num);
}
}
public static void bbTSP() {
isVisited[posStartX][posStartY] = true;
queue.enq(starte);
queue.enq(null);
do{
Element curE = (Element)queue.deq();
while(curE != null) {
addElementToQueue(curE);
curE = (Element)queue.deq();
if(ende.equals(curE)) {
ende = curE;
return ;
}
}
queue.enq(null);
}while(!queue.empty());
}
private static void addElementToQueue(Element e) {
// 對上下左右進行處理 , 沒超出邊界且不是障礙物時 , 加入隊列
// 向左走
if(e.y-1>=0 && maze[e.x][e.y-1] != 'X' && !isVisited[e.x][e.y-1]) {
Element e1 =new Element(e.x,e.y-1,e.num+1,e);
queue.enq(e1);
isVisited[e.x][e.y-1] = true;
e1.print();
}
// 向右走
if(e.y+1 <n && maze[e.x][e.y+1] != 'X' &&!isVisited[e.x][e.y+1]) {
Element e1 =new Element(e.x,e.y+1,e.num+1,e);
queue.enq(e1);
isVisited[e.x][e.y+1] = true;
e1.print();
}
// 向上走
if(e.x-1 >0 && maze[e.x-1][e.y] != 'X' && !isVisited[e.x-1][e.y]) {
Element e1 =new Element(e.x-1,e.y,e.num+1,e);
queue.enq(e1);
isVisited[e.x-1][e.y] = true;
e1.print();
}
// 向下走
if(e.x+1 <n && maze[e.x+1][e.y] != 'X' && !isVisited[e.x+1][e.y]) {
Element e1 =new Element(e.x+1,e.y,e.num+1,e);
queue.enq(e1);
isVisited[e.x+1][e.y] = true;
e1.print();
}
}
public static void main(String args[]) {
new MiGong();
bbTSP();
System.out.println(ende.num);
}
}
// 下面建一個 Queue 類
class Queue extends java.util.Vector {
public Queue() {
super();
}
public synchronized void enq(Object x) {
super.addElement(x);
}
public synchronized Object deq() {
/* 隊列若爲空，引發 EmptyQueueException 異常 */
if (this.empty())
throw new EmptyQueueException();
Object x = super.elementAt(0);
super.removeElementAt(0);
return x;
}
public synchronized Object front() {
if (this.empty())
throw new EmptyQueueException();
return super.elementAt(0);
}
public boolean empty() {
return super.isEmpty();
}
public synchronized void clear() {
super.removeAllElements();
}
public int search(Object x) {
return super.indexOf(x);
}
}
class EmptyQueueException extends java.lang.RuntimeException {
public EmptyQueueException() {
super();
}
}
四． 動態規劃算法。
多邊形問題：（ java ）
import java.util.Scanner;
import java.util.StringTokenizer;
public class PolygonGame {
int[] v;
char[] op;
int arr[][][];
int n ;
int minf,maxf;
void input() {
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
arr = new int[n+1][n+1][2];
v = new int[n+1];
op = new char[n+1];
sc.nextLine();
String inf = sc.nextLine();
sc.close();
StringTokenizer st = new StringTokenizer(inf);
for(int i = n;i>=1; i--) {
v[i] = Integer.parseInt(st.nextToken());
op[i] = st.nextToken().charAt(0);
}
for(int i = 1; i<=n; i++) {
arr[i][1][0] = arr[i][1][1] = v[i];
}
}
void minMax(int i,int s,int j) {
int a = arr[i][s][0];
int b = arr[i][s][1];
int r = (i+s-1)%n+1;
int c = arr[r][j-s][0];
int d = arr[r][j-s][1];
if(op[r] == '+') {// 當運算符爲加時
minf = a + c;
maxf = b + d;
}
else { // 當運算符爲乘時
int[]e = new int[5];
e[1] = a * c;
e[2] = a * d;
e[3] = b * c;
e[4] = b * d;
minf = Math.min(Math.min(e[1],e[2]),Math.min(e[3],e[4]));
maxf = Math.max(Math.max(e[1],e[2]),Math.max(e[3],e[4]));
}
}
int polyMax() {
for(int j = 2;j<=n;j++) { // 鏈的結束位置
for(int i = 1;i<=n;i++) { // 鏈的起始位置
for(int s = 1;s<j; s++) { // 把一條鏈斷開的位置有 j-1 種情況，因爲第一步斷開了一條邊
minMax(i,s,j);
if(arr[i][j][0] > minf) arr[i][j][0] = minf; // 更新值操作
if(arr[i][j][1] < maxf) arr[i][j][1] = maxf;
}
}
}
int temp = 0; // 求出從不同伴置斷開的最大值勤
for(int i = 1; i<= n; i++) {
if(temp < arr[i][n][1]) temp = arr[i][n][1];
}
System.out.println(temp);
return temp;
}
public static void main(String args[]) {
PolygonGame pg = new PolygonGame();
pg.input();
pg.polyMax();
}
}
五． 貪心算法。
Dijkstra- 單源最短路徑(java)
import java.util.Scanner;
public class Dijkstra {
private static int n ;
private static int[][] route; // 路徑長度
private static int[] res; // 最終結果
private static boolean[] hasDone; // 已解決的問題序列
private static boolean[] notDone; // 未解決的問題序列
/* 利用 Dijkstra 算法求解從頂點號 0 到其它頂點的最短路徑
* 最終結果保存在了 res 數組序列中
*/
public Dijkstra() {
try {
Scanner sc = new Scanner(System.in);
// 初始化問題的規模
n = sc.nextInt();
route = new int[n][n];
res = new int[n];
hasDone = new boolean[n];
notDone = new boolean[n];
notDone[0] = false;
hasDone[0] = true;
sc.nextLine();
String inf = "";
String[] dataArr;
for(int i = 0;i<n;i++) {
inf = sc.nextLine();
dataArr = inf.split(" ");
for(int j = 0;j<n;j++) {
route[i][j] = Integer.parseInt(dataArr[j]);
}
}
/*System.out.println(" 問題規模 " + n );
for(int i = 0;i<n;i++) {
for(int j = 0;j<n;j++) {
System.out.print(route[i][j] + "/t");
}
System.out.println();
}*/
for(int i = 1; i<n;i++) {
res[i] = route[0][i];
notDone[i] = true;
hasDone[i] = false;
}
sc.close();
}catch(Exception e) {
}
}
public static void dijkstra() {
// System.out.println(" 進行求解 ");
// printMinRoute();
int i = 1; // 問題求解完成數 ( 一共要完成 n 個 )
//int minN = 0; // 當前要處理的頂點
while(i<n) { // 只要還沒有處理完 n 個數據
update(getMinRoute());
// printMinRoute();
i++;
}
// printMinRoute();
for(i =1;i<n-1;i++) {
System.out.print(res[i] + " ");
}
System.out.println(res[n-1]);
}
private static int getMinRoute() {
int minN = -1;
int temp = Integer.MAX_VALUE;
for(int i = 1; i< n; ++i) {
/*for(int t =1;t<n;t++) {
System.out.print(res[t] + " ");
}
System.out.println()*/;
if(notDone[i] == true && res[i] > 0 ) {
if(temp > res[i]) {
temp = res[i];
minN = i;
}
}
}
return minN;
}
/* 輸出最後結果
*
*/
public static void printMinRoute() {
System.out.println(" 打印結果 ");
for(int i = 1; i<n ;i++) {
System.out.println(" 到第 " + i + " 個頂點的最小路徑爲 " + res[i]);
}
}
/*
* 對由於加入頂點 i 而引起的數據變化 , 進行更新操作
*/
private static void update(int i) {
// 結果序列進行更新
for(int k = 1 ; k< n ; k++) {
if(i!=k && notDone[k] == true && route[i][k]>0) {
if(route[i][k] + res[i] < res[k] || res[k] <0) {
res[k] = route[i][k] + res[i];
}
}
}
hasDone[i] = true;
// 從未解決序列中刪除
notDone[i] = false;
}
public static void main(String args[]) {
Dijkstra ds = new Dijkstra();
dijkstra();
}
}
六． 近似算法。
具有三角不等式特點的旅行售貨員問題(C).
#include <stdio.h>
//#define _DEBUG
#undef _DEBUG
// 定義最大值宏
#define M 9999
// 定義規模
#define SIZE 25
// 定義初始化數據
#define INIT -1
// 定義一個 25*25 的二維數組
int arr[SIZE][SIZE];
// 最小消費向量
// 最小集
int closest[SIZE];
// 訪問標誌數組
int lowcost[SIZE];
bool visited[SIZE];
// 加入頂點的順序
int addorder[SIZE];
// 初始化操作
void init()
{
int i,j;
for(i=0;i<SIZE;++i)
{
for(j=0;j<SIZE;++j)
{
arr[i][j] = M;
}
lowcost[i] = INIT;
closest[i] = INIT;
visited[i] = false;
addorder[i] = INIT;
}
}
// 輸出矩陣
void printArray(int num)
{
int i,j;
for(i=1;i<=num;++i)
{
for(j=1;j<=num;++j)
{
printf("%d/t",arr[i][j]);
}
printf("/n");
}
}
// 輸出最小生成樹
void printMin(int num)
{
#ifdef _DEBUG
printf(" 最小生成樹 /n");
printf("4 的值 %d 值 %d/n",closest[4],lowcost[4]);
#endif
int i;
for(i=1;i<num;i++)
{
printf("%d %d/n",closest[addorder[i]],addorder[i]);
}
}
//prime 算法
void prime(int num)
{
// 第一個頂點初始化
visited[1] = true;
// 初始化最小消費和最小集合
for(int i = 2;i<=num;i++)
{
lowcost[i] = arr[1][i];
closest[i] = 1;
visited[i] = false;
}
//num-1 次循環完後 , 解決問題 .
for(i = 1;i<num;i++)
{
int min = M;
int j = 1;
for(int k = 2;k<=num;k++)
{
if((lowcost[k]<min) && (!visited[k]))
{
min = lowcost[k];
j = k;
}
}
visited[j] = true;
addorder[i] = j;
#ifdef _DEBUG
printf(" 第 %d 次加入了頂點 %d/n",i,j);
#endif
for(k = 1;k<=num;k++)
{
if((arr[j][k]<lowcost[k]) && (!visited[k]))
{
lowcost[k] = arr[j][k];
closest[k] = j;
}
}
#ifdef _DEBUG
printMin(num);
#endif
}
}
// 前序遍歷輸出結果
void printMinValue(int num)
{
int verarr[SIZE];
// 開始頂點 1
verarr[1] = 1;
bool s[SIZE][SIZE];
int j,k;
for(j=1;j<=num;++j)
{
for(k=1;k<=num;++k)
{
s[j][k] = false;
}
}
bool findflag = false;
int i = 1;
// 定義一個 verarr 向量位置的變量，同時可做爲結束循環的標誌。
int verpos = 1;
// 進行 num-2 次循環
while(i>0)
{
// 重置 findflag;
findflag = false;
// 在 addorder 找 verarr[i-1] 的第一個鄰接頂點
for(j=1;j<num;++j)
{
if(closest[addorder[j]] == verarr[i] && !s[verarr[i]][addorder[j]])
{
s[verarr[i]][addorder[verpos]] = true;
verarr[++verpos] = addorder[j];
findflag = true;
++i;
break;
}
}
// printf("i 的值 %d/n",i);
if(verpos == num)
break;
if(!findflag)
{
i--;
}
}
int sum = 0;
for(k=1;k<num;k++)
{
#ifdef _DEBUG
printf("%d/t",verarr[k]);
printf("%d/t %d/t %d/n",verarr[k],verarr[k+1],arr[verarr[k]][verarr[k+1]]);
#endif
sum = sum + arr[verarr[k]][verarr[k+1]];
}
sum += arr[verarr[k]][1];
printf("%d/n",sum);
}
// 主方法
int main()
{
#ifdef _DEBUG
printf(" 調試模式 /n");
#endif
// 執行初始化操作
init();
// 頂點數和邊數
int vertex_num,edge_num;
// 輸入頂點數
scanf("%d",&vertex_num);
// 輸入邊數
scanf("%d",&edge_num);
// 輸入值
int loop;
// 矩陣從索引 1 開始
for(loop = 1;loop<=edge_num;++loop)
{
int tmpver1,tmpver2,tmpedg;
scanf("%d%d%d",&tmpver1,&tmpver2,&tmpedg);
// 存入矩陣 ( 記住是無向圖 )
arr[tmpver1][tmpver2] = tmpedg;
arr[tmpver2][tmpver1] = tmpedg;
}
prime(vertex_num);
#ifdef _DEBUG
printArray(vertex_num);
#endif
// printMin(vertex_num);
printMinValue(vertex_num);
return 0;
}