CodeForces

Berland shop sells $n$

Petya knows that he needs all three types of vitamins to stay healthy. What is the minimum total price of juices that Petya has to buy to obtain all three vitamins? Petya obtains some vitamin if he buys at least one juice containing it and drinks it.

Input

The first line contains a single integer $n$

Each of the next $n$

Output

Print -1 if there is no way to obtain all three vitamins. Otherwise print the minimum total price of juices that Petya has to buy to obtain all three vitamins.

Examples

Input

4
5 C
6 B
16 BAC
4 A

Output

Input

2
10 AB
15 BA

Output

-1

Input

5
10 A
9 BC
11 CA
4 A
5 B

Output

Input

6
100 A
355 BCA
150 BC
160 AC
180 B
190 CA

Output

Input

2
5 BA
11 CB

Output

Note

In the first example Petya buys the first, the second and the fourth juice. He spends $5 + 6 + 4 = 15$

In the second example Petya can't obtain all three vitamins, as no juice contains vitamin "C".

你猜我的代码错在哪了，找找看？

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<iostream>
using namespace std;
typedef long long  LL;
const double pi=acos(-1.0);
const double e=exp(1);
const int N = 100009;


int main()
{
    char kind[5];
    LL i,p,j;
    LL n,x;
    LL a,b,c,ab,ac,bc,abc;
    LL flag=0;
    a=b=c=ab=ac=bc=abc=100009;
    scanf("%lld",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%lld %s",&x,kind);
        if((strcmp(kind,"BA")==0)||(strcmp(kind,"AB")==0))
        {
            if(x<ab)
                ab=x;
        }
        else if((strcmp(kind,"AC")==0)||(strcmp(kind,"CA")==0))
        {
            if(x<ac)
                ac=x;
        }
        else if((strcmp(kind,"BC")==0)||(strcmp(kind,"CB")==0))
        {
            if(x<bc)
                bc=x;
        }
        else if(kind[0]=='A')
        {
            flag++;
            if(x<a)
                a=x;
        }
        else if(kind[0]=='B')
        {
            flag++;
            if(x<b)
                b=x;
        }
        else if(kind[0]=='C')
        {
            flag++;
            if(x<c)
                c=x;
        }
        else
        {
            if(x<abc)
                abc=x;
        }
        kind[0]=0;
    }

    LL head=99999999999;
    if(a<100009&&b<100009&&c<100009)
        head=a+b+c;
    if(bc<100009&&a<100009&&(a+bc<head))
        head=a+bc;
    if(abc<100009&&a<100009&&(a+abc<head))
        head=a+abc;
    if(abc<100009&&b<100009&&(b+abc<head))
        head=b+abc;
    if(abc<100009&&c<100009&&(c+abc<head))
        head=c+abc;
    if(ac<100009&&b<100009&&(b+ac<head))
        head=b+ac;
    if(ab<100009&&c<100009&&(c+ab<head))
        head=c+ab;
    if(ac<100009&&ab<100009&&(ac+ab<head))
        head=ac+ab;
    if(ab<100009&&bc<100009&&(ab+bc<head))
        head=ab+bc;
    if(ac<100009&&bc<100009&&(ac+bc<head))
        head=ac+bc;
    if(abc<100009&&abc<head)
        head=abc;

    if(head!=99999999999)
        printf("%lld
",head);
    else
        printf("-1
");
    return 0;

}

if else if else if if if .......因为这种问题错过几遍了，心里没有点Number B 吗？！！

明明"ABC"的串还没有比过，你就直接比较kind[0]??? 脑子呢？？？

写if else 的时候能不能心里想着逻辑点，能不能走点心？

正确代码：

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cstring>
  4 #include<string>
  5 #include<cmath>
  6 #include<algorithm>
  7 #include<queue>
  8 #include<stack>
  9 #include<deque>
 10 #include<map>
 11 #include<iostream>
 12 using namespace std;
 13 typedef long long  LL;
 14 const double pi=acos(-1.0);
 15 const double e=exp(1);
 16 const int N = 100009;
 17 
 18 
 19 int main()
 20 {
 21     char kind[5];
 22     LL i,p,j;
 23     LL n,x;
 24     LL a,b,c,ab,ac,bc,abc;
 25     LL flag=0;
 26     a=b=c=ab=ac=bc=abc=100009;
 27     scanf("%lld",&n);
 28     for(i=1;i<=n;i++)
 29     {
 30         scanf("%lld %s",&x,kind);
 31         if((strcmp(kind,"BA")==0)||(strcmp(kind,"AB")==0))
 32         {
 33             if(x<ab)
 34                 ab=x;
 35         }
 36         else if((strcmp(kind,"AC")==0)||(strcmp(kind,"CA")==0))
 37         {
 38             if(x<ac)
 39                 ac=x;
 40         }
 41         else if((strcmp(kind,"BC")==0)||(strcmp(kind,"CB")==0))
 42         {
 43             if(x<bc)
 44                 bc=x;
 45         }
 46         else if(strcmp(kind,"A")==0)
 47         {
 48             flag++;
 49             if(x<a)
 50                 a=x;
 51         }
 52         else if(strcmp(kind,"B")==0)
 53         {
 54             flag++;
 55             if(x<b)
 56                 b=x;
 57         }
 58         else if(strcmp(kind,"C")==0)
 59         {
 60             flag++;
 61             if(x<c)
 62                 c=x;
 63         }
 64         else
 65         {
 66             if(x<abc)
 67                 abc=x;
 68         }
 69         kind[0]=0;
 70     }
 71 
 72     LL head=99999999999;
 73     if(a<100009&&b<100009&&c<100009)
 74         head=a+b+c;
 75     if(bc<100009&&a<100009&&(a+bc<head))
 76         head=a+bc;
 77     if(abc<100009&&a<100009&&(a+abc<head))
 78         head=a+abc;
 79     if(abc<100009&&b<100009&&(b+abc<head))
 80         head=b+abc;
 81     if(abc<100009&&c<100009&&(c+abc<head))
 82         head=c+abc;
 83     if(ac<100009&&b<100009&&(b+ac<head))
 84         head=b+ac;
 85     if(ab<100009&&c<100009&&(c+ab<head))
 86         head=c+ab;
 87     if(ac<100009&&ab<100009&&(ac+ab<head))
 88         head=ac+ab;
 89     if(ab<100009&&bc<100009&&(ab+bc<head))
 90         head=ab+bc;
 91     if(ac<100009&&bc<100009&&(ac+bc<head))
 92         head=ac+bc;
 93     if(abc<100009&&abc<head)
 94         head=abc;
 95 
 96     if(head!=99999999999)
 97         printf("%lld
",head);
 98     else
 99         printf("-1
");
100     return 0;
101 
102 }

View Code

相关阅读:
ElasticSearch6学习(1)-安装Elasticsearch
Ubuntu 18.04 安装java8
windows10 php7安装mongodb 扩展
https加密解密过程详解
Beanstalkd，zeromq，rabbitmq的区别
PHP中的++和--
win10 git bash 闪退
谈下WebSocket介绍，与Socket的区别
Bridge桥接模式(结构型模式)
Apater适配器模式(结构型模式)

原文地址：https://www.cnblogs.com/daybreaking/p/9740277.html