CodeForces - 1042B

Berland shop sells nn kinds of juices. Each juice has its price cici. Each juice includes some set of vitamins in it. There are three types of vitamins: vitamin "A", vitamin "B" and vitamin "C". Each juice can contain one, two or all three types of vitamins in it.

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 nn (1≤n≤1000)(1≤n≤1000) — the number of juices.

Each of the next nn lines contains an integer cici (1≤ci≤100000)(1≤ci≤100000) and a string sisi — the price of the ii-th juice and the vitamins it contains. String sisi contains from 11 to 33 characters, and the only possible characters are "A", "B" and "C". It is guaranteed that each letter appears no more than once in each string sisi. The order of letters in strings sisi is arbitrary.

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
15
Input
2
10 AB
15 BA
Output
-1
Input
5
10 A
9 BC
11 CA
4 A
5 B
Output
13
Input
6
100 A
355 BCA
150 BC
160 AC
180 B
190 CA
Output
250
Input
2
5 BA
11 CB
Output
16

Note

In the first example Petya buys the first, the second and the fourth juice. He spends 5+6+4=155+6+4=15 and obtains all three vitamins. He can also buy just the third juice and obtain three vitamins, but its cost is 1616, which isn't optimal.

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\n",head);
    else
        printf("-1\n");
    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\n",head);
 98     else
 99         printf("-1\n");
100     return 0;
101 
102 }
View Code

 

posted @ 2018-10-03 17:36  Daybreaking  阅读(348)  评论(0)  编辑  收藏  举报