参数目录化，就是将 类似 http://www.abc.com/store/store.aspx?id=1024 这样的网址，对外改为 http://www.abc.com/1024。

要实现这种功能，可以用以下三种方式（应该不仅限这三种吧！）

1、用微软的 Microsoft URL 重写模块 2.0，但只能给IIS7使用，IIS6不行。

64位：

http://www.microsoft.com/downloads/zh-cn/details.aspx?familyid=1b8c7bd8-8824-4408-b8fc-49dc7f951a00

32位：

http://www.microsoft.com/zh-cn/download/details.aspx?id=5747

2、isapi_rewrite

http://www.helicontech.com/download-isapi_rewrite3.htm

但完全版有日期限制，如果不想掏钱，有大牛破解了，可以用下面这个：

http://download.csdn.net/detail/keke0307/3867086

3、urlrewriter.net

这个有源代码，可以自己编译。

机缘巧合之下，我用了方法一 和 二。因为开发机器是WIN7，装了IIS7，所以用微软的重写模块；而服务器是WIN2003，就用了isapi_rewrite。

下面就这两种方法分别做一点心得介绍。

事实上，这两种方法大同小异，IIS基本上不用怎么配置，没有网上说的那么玄妙，又是勾选，又是映射，又是权限之类，关键在于要写对正则表达式。

1、微软的 Microsoft URL 重写模块 2.0

安装好之后，就可以改写网站下的web.config了。

<configuration>
  <system.web>
    <compilation debug="true" targetFramework="4.0"/>
    <httpRuntime/>
  </system.web>
  <system.webServer>
    <rewrite>
      <rules>
	<!-- 实现http://localhost/1024 ==> http://localhost/store/store.aspx?id=1024 -->
        <rule name="storecode">
          <match url="^([1-9][0-9]*)/?$" ignoreCase="true"/>
          <action type="Rewrite" url="/store/store.aspx?id={R:1}"/>
        </rule>
	<!-- 实现http://localhost/1024/p=1&c=1 ==> http://localhost/store/store.aspx?id=1024&p=1&c=1 -->
        <rule name="storecode with param">
          <match url="^([1-9][0-9]*)/([^/]+)/?$" ignoreCase="true"/>
          <action type="Rewrite" url="/store/store.aspx?id={R:1}&{R:2}"/>
        </rule>
	<!-- 下面这些据说是从Discuz论坛里导出来的，肯定是正确的，供学习借鉴 -->
        <!--
                <rule name="已导入的规则 1">
                    <match url="^userinfo-([0-9]+)\.html$" ignoreCase="false" />
                    <action type="Rewrite" url="userinfo.aspx?userid={R:1}" 

appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 2">
                    <match url="^showforum-([0-9]+)\.html$" ignoreCase="false" />
                    <action type="Rewrite" url="showforum.aspx?forumid={R:1}" 

appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 3">
                    <match url="^showtopic-([0-9]+)\.html$" ignoreCase="false" />
                    <action type="Rewrite" url="showtopic.aspx?topicid={R:1}" 

appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 4">
                    <match url="^showforum-([0-9]+)-([0-9]+)\.html$" ignoreCase="false" />
                    <action type="Rewrite" url="showforum.aspx?forumid={R:1}&page=

{R:2}" appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 5">
                    <match url="^showtopic-([0-9]+)-([0-9]+)\.html$" ignoreCase="false" />
                    <action type="Rewrite" url="showtopic.aspx?topicid={R:1}&page=

{R:2}" appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 6">
                    <match url="^archiver/showforum-([0-9]+)\.html$" ignoreCase="false" />
                    <action type="Rewrite" url="archiver/showforum.aspx?forumid={R:1}" 

appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 7">
                    <match url="^archiver/showtopic-([0-9]+)\.html$" ignoreCase="false" />
                    <action type="Rewrite" url="archiver/showtopic.aspx?topicid={R:1}" 

appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 8">
                    <match url="^archiver/showtopic-([0-9]+)-([0-9]+)\.html$" 

ignoreCase="false" />
                    <action type="Rewrite" url="archiver/showtopic.aspx?topicid={R:1}

&page={R:2}" appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 9">
                    <match url="^archiver/showforum-([0-9]+)-([0-9]+)\.html$" 

ignoreCase="false" />
                    <action type="Rewrite" url="archiver/showforum.aspx?forumid={R:1}

&page={R:2}" appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 10">
                    <match url="^tools/rss-([0-9]+)\.html$" ignoreCase="false" />
                    <action type="Rewrite" url="tools/rss.aspx?forumid={R:1}" 

appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 11">
                    <match url="^tools/spacerss-([0-9]+)\.html$" ignoreCase="false" />
                    <action type="Rewrite" url="tools/rss.aspx?uid={R:1}&type=space" 

appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 12">
                    <match url="^tools/photorss-([0-9]+)\.html$" ignoreCase="false" />
                    <action type="Rewrite" url="tools/rss.aspx?uid={R:1}&type=photo" 

appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 13">
                    <match url="^space\/((\w|\s)+)((\/?))?$" ignoreCase="false" />
                    <action type="Rewrite" url="space/index.aspx?user={R:1}" 

appendQueryString="false" />
                </rule>
                <rule name="已导入的规则 14">
                    <match url="^space\/((\w|\s|-)+)((\/?))?\?((.*)+)$" ignoreCase="false" 

/>
                    <action type="Rewrite" url="space/index.asp      

      
      
    

      

Given an NxM (N rows and M columns) integer matrix with non-negative values (0..MAX_INT inclusive). What is the maximum sum from going top left (0, 0) to bottom right (N-1, M-1) ? The condition is that when you're at point (p, q), you can only move to either right (p, q+1) or down (p+1, q).

Expected time complexity O(N*M)
Expected space complexity O(N+M) 

about space complexity, need elaborate on it.

package com.zhuyu_deng.test;



public class Test
{
	private static int findOptPath(int[][] a)
	{
		int d[][] = new int[a.length][a[0].length];
		int m = a.length;
		int n = a[0].length;
		
		d[0][0] = a[0][0];
		for (int i = 1; i < m; ++i)
			d[i][0] = d[i - 1][0] + a[i][0];
		for (int j = 1; j < n; ++j)
			d[0][j] = d[0][j - 1] + a[0][j];
		
		for (int i = 1; i < m; ++i)
		{
			for (int j = 1; j < n; ++j)
			{

				if (d[i-1][j] > d[i][j-1])
					d[i][j] = a[i][j] + d[i-1][j];
				else
					d[i][j] = a[i][j] + d[i][j-1];
			}
		}
		
		for (int i = 0; i < m; ++i)
		{
			for (int j = 0; j < n; ++j)
				System.out.print(a[i][j] + "  ");
			System.out.println();
		}
		System.out.println();
		
		for (int i = 0; i < m; ++i)
		{
			for (int j = 0; j < n; ++j)
				System.out.print(d[i][j] + "  ");
			System.out.println();
		}
		return d[m - 1][n - 1];
	}

	public static void main(String args[])
	{
		// int[] a = {-2,11,-4,13,-5,-2};
		int[][] b = { { 0, -2, -7, 0 }, { 9, 2, -6, 2 }, { -4, 1, -4, 1 },
				{ -1, 8, 0, -2 } };
		int [][] matrix = 
			{	{2,3,4,1},
				{1,1,3,9},
				{2,2,3,1},
				{2,2,3,1}
			};
		System.out.println(findOptPath(matrix));
	}
}