<h1 id="heading">题目</h1>
<p>{% raw %}</p>
<div><div class="fold_hider"><div class="close hider_title">点击显/隐题目</div></div><div class="fold">
    <div class="oj">   
        <div class="part" title="Description">
{% endraw %}
Finally term begins. luras loves school so much as she could skip the class happily again.(wtf?)  
<p>Luras will take n lessons in sequence(in another word, to have a chance to skip xDDDD).</p>
<p>For every lesson, it has its own type and value to skip.</p>
<p>But the only thing to note here is that luras can't skip the same type lesson more than twice.</p>
<p>Which means if she have escaped the class type twice, she has to take all other lessons of this type.</p>
<p>Now please answer the highest value luras can earn if she choose in the best way.</p>
<p>{% raw %}<br>
</div><br>
<div class="part" title="Input"><br>
{% endraw %}</p>
<p>The first line is an integer T which indicates the case number.</p>
<p>And as for each case, the first line is an integer n which indicates the number of lessons luras will take in sequence.</p>
<p>Then there are n lines, for each line, there is a string consists of letters from 'a' to 'z' which is within the length of 10,<br>
and there is also an integer which is the value of this lesson.</p>
<p>The string indicates the lesson type and the same string stands for the same lesson type.</p>
<p>It is guaranteed that——</p>
<p>T is about 1000</p>
<p>For 100% cases, 1 &lt;= n &lt;= 100，1 &lt;= |s| &lt;= 10, 1 &lt;= v &lt;= 1000</p>
<p>{% raw %}<br>
</div><br>
<div class="part" title="Output"><br>
{% endraw %}</p>
<p>As for each case, you need to output a single line.<br>
there should be 1 integer in the line which represents the highest value luras can earn if she choose in the best way.</p>
<p>{% raw %}<br>
</div><br>
<div class="samp"><br>
<div class="clear"></div><br>
<div class="input part" title="Sample Input"><br>
{% endraw %}</p>
<p>2<br>
5<br>
english 1<br>
english 2<br>
english 3<br>
math 10<br>
cook 100<br>
2<br>
a 1<br>
a 2</p>
<p>{% raw %}<br>
</div><br>
<div class="output part" title="Sample Output"><br>
{% endraw %}</p>
<p>115<br>
3</p>
<p>{% raw %}<br>
</div><br>
<div class="clear"></div><br>
</div><br>
</div></p>
</div></div>
{% endraw %}

<h1 id="heading-1">题解</h1>
<p>跳过一定数目的课程,使最后得到的总值最多<br>
同一门课最多跳过两次</p>
<p>因此,将每一门课 value 最高的两门选出来加起来即可</p>
<p>用一个结构体维护所有的课程,根据课程名称和 value 排序即可</p>
<h1 id="heading-2">代码</h1>
<div><div class="fold_hider"><div class="close hider_title">点击显/隐代码</div></div><div class="fold">```cpp Skip the Class https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
/*/
#define debug
#include <ctime>
//*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
<p>const int maxn = 105;</p>
<p>struct Node{<br>
string name;<br>
int value;<br>
bool operator &lt; (const Node rhs)const{<br>
if(name == rhs.name)<br>
return value &gt; rhs.value;<br>
return name &lt; rhs.name;<br>
}<br>
};</p>
<p>Node node[maxn];</p>
<p>int main(){<br>
#ifdef debug<br>
freopen(&quot;in.txt&quot;, &quot;r&quot;, stdin);<br>
int START = clock();<br>
#endif<br>
cin.tie(0);<br>
cin.sync_with_stdio(false);</p>
<pre><code>int T;
cin &gt;&gt; T;
while(T--){
    int n;
    cin &gt;&gt; n;
    for(int i=0;i&lt;n;i++)
        cin &gt;&gt; node[i].name &gt;&gt; node[i].value;
    sort(node,node+n);

    string ls = &quot;&quot;;
    int cnt = 0;
    int ans = 0;
    for(int i=0;i&lt;n;i++){
        if(ls == node[i].name){
            if(cnt == 2){
                continue;
            }else{
                cnt++;
                ans += node[i].value;
            }
        }else{
            ls = node[i].name;
            cnt = 1;
            ans += node[i].value;
        }


    }
    cout &lt;&lt; ans&lt;&lt;endl;
}

#ifdef debug
printf(&quot;Time:%.3fs.\n&quot;, double(clock() - START) / CLOCKS_PER_SEC);
#endif
return 0;
</code></pre>
<p>}</p>
<pre style="background-color:#fff">&lt;/div&gt;</pre>

HDU 6015.Skip the Class