CodeForces - 1042B （状压dp）

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 11to 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.

     有n个物品，每个物品有他的价值v，以及她所含的维生素ABC（可能有1种或者2种或者3种），请问怎么买，才能在最小的代价下同时获得ABC三个元素呢？

     有两种办法，一个是在状态这么小的情况下暴力枚举条件找就可以，二是进行状压dp,用dp[i][j]表示到第i个物品为止，获得j状态的最小代价是多少。这里只列出dp方法的代码

#include<bits/stdc++.h>
using namespace std;
int n;
const int inf = 0x3f3f3f3f;
int m[1005][10];
int main(){
    memset(m,inf,sizeof m);
    scanf("%d",&n);
    m[0][0]=0;
    for(int i=1;i<=n;i++){
        for(int j=0;j<8;j++)
            m[i][j]=m[i-1][j];
        int v,kind=0,len;
        char r[10];
        scanf("%d",&v);scanf("%s",r);
        len=strlen(r);
        for(int i=0;i<len;i++){
            int t=r[i]-'A';
            kind+=(1<<t);
        }
        for(int j=0;j<8;j++){
            int t=(kind|j);
            m[i][t]=min(m[i][t],m[i-1][j]+v);
        }
    }
    if(m[n][7]>=inf){
        m[n][7]=-1;
    }
    printf("%d\n",m[n][7]);
}