stl应用（map）或字典树（有点东西）

M - Violet Snow

Gym - 101350M

Every year, an elephant qualifies to the Arab Collegiate Programming Competition. He graduated this year, but that’s irrelephant. What’s important is that the location of the competition might not have been the same every year. Therefore, after every trip, he always has leftover money in the currency of the country he visited.

Now he wants to see how much Jordanian Dinars he has after all those competitions. Can you help him convert the leftover money from all competitions to Jordanian Dinar, if that makes any cents?

Input

The first line of input is T – the number of test cases.

The first line of each test case contains C and N (1 ≤ C, N ≤ 100000), the number of currency types and the number of competitions, respectively.

The next C lines each contain the name of the currency C_i of maximum length 10 in lowercase and/or uppercase letters, and the value V_i of that currency in Jordanian Dinar (0 < V_i ≤ 1000). The names are case-sensitive.

The next N lines each contains an amount left over from each competition (0 ≤ N_i ≤ 1000), and the name of the currency of that amount (it is guaranteed that the name was either given in the input or is “JD”).

Output

For each test case, print on a single line the total amount of money he has in Jordanian Dinar(JD) rounded to 6 decimal digits.

Example

Input

1
3 5
dollar 0.71
euro 0.76
turkish 0.17
5.1 dollar
6 dollar
7 turkish
3 euro
1.1 JD

Output

12.451000
https://blog.csdn.net/sevenjoin/article/details/81943864

自动建立key － value的对应。key 和 value可以是任意你需要的类型。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
#include <map>
//#define PI 3.1415926
using namespace std;
const double PI = acos(-1.0);


int main()
{
    int n ;
    scanf("%d" , &n);
    while(n--)
    {
        map<string , double>ma;
        int m , p ;
        scanf("%d%d" , &m ,&p);
        for(int i = 0 ; i < m ; i++)
        {
            string a ;
            double b ;
            cin >> a ;
            scanf("%lf" , &b);
            //map容器的三种插入，第一种key可以覆盖 ，其他两种不可以
            ma[a] = b;
            ma.insert(map<string , double>::value_type(a , b));
            ma.insert(pair<string , double>(a , b));
        }
        double sum = 0 ;
        for(int i = 0 ; i < p ; i++)
        {
            string a ;
            double b;
            scanf("%lf" , &b);
            cin >> a;
            if(a == "JD")
                sum += b ;
            else
                sum += ma[a] * b ;
        }
        printf("%.6lf
" , sum);
    }

    return 0 ;
}

应该没几个人用字典树吧

字典树是完全可行的，我就是因为没对用过的树初始化所有就wa了，

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
using namespace std;
int tree[500009][65];
int k = 0 ;
double vis[500009] ;
double sum = 0 ;
void winsert(char *a , double val)
{
    int len = strlen(a);
    int p = 0 ;
    for(int i = 0 ; i < len ; i++)
    {
        int c = a[i] - 'A';
        if(!tree[p][c])
            tree[p][c] = ++k ;
        p = tree[p][c];
    }
    vis[p] = val ;
}

double wsearch(char *a)
{
    int p = 0 ;
    int len = strlen(a);
    for(int i = 0 ; i < len ; i++)
    {
        int c = a[i] - 'A';
        if(!tree[p][c]) return 1 ;
        p = tree[p][c];
    }
    if(vis[p])
        return vis[p];
}


int main()
{
    int n ;
    scanf("%d" , &n);
    while(n--)
    {
        k = 0 ;
        memset(vis, 0 , sizeof(vis));
        memset(tree , 0 , sizeof(tree));
        sum = 0 ;
        int m , p ;
        scanf("%d%d" , &m ,&p);
       // cout << m << " " << p << endl ;
        char s[100];
        double val ;
        for(int i = 0 ; i < m ; i++)
        {
            scanf("%s%lf" ,s , &val);
            winsert(s , val);
        }
        for(int i = 0 ; i < p ; i++)
        {
            scanf("%lf%s" , &val , s);
            sum += val * wsearch(s);
        }
        printf("%.6lf
" , sum);

    }



    return 0 ;
}