Finally term begins. luras loves school so much as she
could skip the class happily again.(wtf?)
Luras will take n lessons in
sequence(in another word, to have a chance to skip xDDDD).
For every
lesson, it has its own type and value to skip.
But the only thing to note
here is that luras can‘t skip the same type lesson more than twice.
Which
means if she have escaped the class type twice, she has to take all other
lessons of this type.
Now please answer the highest value luras can earn
if she choose in the best way.
The first line is an integer T which indicates the case
number.
And as for each case, the first line is an integer n which
indicates the number of lessons luras will take in sequence.
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,
and there is also an integer which is
the value of this lesson.
The string indicates the lesson type and the
same string stands for the same lesson type.
It is guaranteed
that——
T is about 1000
For 100% cases, 1 <= n <= 100，1 <=
|s| <= 10, 1 <= v <= 1000
As for each case, you need to output a single
line.
there should be 1 integer in the line which represents the highest
value luras can earn if she choose in the best way.
2
5
english 1
english 2
english 3
math 10
cook 100
2
a 1
a 2
115
3
http://acm.hdu.edu.cn/showproblem.php?pid=6015
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<list>
#include<vector>
#include<stack>
#include<queue>
#define ll long long
#define inf 0x3f3f3f3f;
using namespace std;
struct node{
char cl[103];
int val;
}nn[103];
int t,n,sum;
bool cmp(node a,node b){
return a.val>b.val;
}
map<string,int>m;
int main(){
scanf("%d",&t);
while(t--){
scanf("%d",&n);
sum=0;
m.clear();
for(int i=0;i<n;i++){
scanf("%s%d",&nn[i].cl,&nn[i].val);
m[nn[i].cl]=0;
}
sort(nn,nn+n,cmp);
for(int i=0;i<n;i++){
if(m[nn[i].cl]<2){
sum+=nn[i].val;
m[nn[i].cl]++;
}
}
printf("%d\n",sum);
}
return 0;
}
