hiho一下 第六十二周 题目1 : Browser Caching stl 应用

题目1 : Browser Caching

时间限制: 10000ms

单点时限: 1000ms

内存限制: 256MB

描述

When you browse the Internet, browser usually caches some documents to reduce the time cost of fetching them from remote servers. Let's consider a simplified caching problem. Assume the size of browser's cache can store M pages. When user visits some URL, browser will search it in the cache first. If the page is already cached browser will fetch it from the cache, otherwise browser will fetch it from the Internet and store it in the cache. When the cache is full and browser need to store a new page, the least recently visited page will be discarded.

Now, given a user's browsing history please tell us where did browser fetch the pages, from the cache or the Internet? At the beginning browser's cache is empty.

输入

Line 1: Two integers N(1 <= N <= 20000) and M(1 <= M <= 5000). N is the number of pages visited and M is the cache size.

Line 2~N+1: Each line contains a string consisting of no more than 30 lower letters, digits and dots('.') which is the URL of the page. Different URLs always lead to different pages. For example www.bing.com and bing.com are considered as different pages by browser.

输出

Line 1~N: For each URL in the input, output "Cache" or "Internet".

提示

Pages in the cache before visiting 1st URL [null, null]

Pages in the cache before visiting 2nd URL [www.bing.com(1), null]

Pages in the cache before visiting 3rd URL [www.bing.com(1), www.microsoft.com(2)]

Pages in the cache before visiting 4th URL [www.bing.com(1), www.microsoft.com(3)]

Pages in the cache before visiting 5th URL [windows.microsoft.com(4), www.microsoft.com(3)]

The number in parentheses is the last visiting timestamp of the page.

样例输入

5 2
www.bing.com
www.microsoft.com
www.microsoft.com
windows.microsoft.com
www.bing.com

样例输出

Internet
Internet
Cache
Internet
Internet

题意，给出一个浏览器，可以缓存m条，如果缓存了，就直接从缓存中读取，否则，从网上下载，如果达到了m条，按最近没有使用的页面淘汰，也就是最近使用的时间最现大最长的淘汰。

stl 简单应用。

直接用map维护所有的缓存，如果存在就输出缓存。再用map记录，最近使用的时间，总的复杂度为o(n * log(n));

#define N 205
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,m;
char str[100];
map<string,int> strm;
map<int,string> timem;
map<int,string>::iterator it;
int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
     while(S2(n,m)!=EOF)
    {
        strm.clear();
        timem.clear();
        FI(n){
            SS(str);
            if(strm.count(str)){
                printf("Cache\n");
                int t = strm[str];
                timem.erase(t);
                strm[str] = i;
                timem[i] = str;
            }
            else{
                printf("Internet\n");
                if(timem.size() >= m){
                    it = timem.begin();
                    strm.erase(it->second);
                    timem.erase(it);
                    strm[str] = i;
                    timem[i] = str;
                }
                else {
                    strm[str] = i;
                    timem[i] = str;
                }
            }
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}