Codeforces Round #510 (Div. 2) B. Vitamins（暴力状压DP）

地址：http://codeforces.com/contest/1042/problem/B

他的情况只有A，B，C，AB，BC，AC，ABC七种，可以先将n个果汁进行两两组合更新七种情况的值，然后在七种两辆组合更新，这是比较暴力的解法

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int N = 1005;

int change(string s)
{
    sort(s.begin(),s.end());
    if(s == "A") return 1;
    if(s == "B") return 2;
    if(s == "C") return 4;
    if(s == "AB") return 3;
    if(s == "BC") return 6;
    if(s == "AC") return 5;
    if(s == "ABC") return 7;
}

int a[15];
int b[N];
int c[N];

int main()
{
    int n;
    scanf("%d",&n);
    memset(a,inf,sizeof(a));
    for(int i = 0;i < n;++i)
    {
        string s;
        cin >> b[i] >> s;
        int x = change(s);
        a[x] = min(a[x],b[i]);
        c[i] = x;
    }
    for(int i = 0;i < n;++i)
    {
        for(int j = i + 1;j < n;++j)
        {
            int x = c[i] | c[j];
            a[x] = min(a[x],b[i] + b[j]);
        }
    }
    for(int i = 1;i <= 7;++i)
    {
        for(int j = i + 1;j <= 7;++j)
        {
            int x = i | j;
            a[x] = min(a[x],a[i] + a[j]);
        }
    }
    if(a[7] == inf){
        printf("-1\n");
    }else{
        printf("%d\n",a[7]);
    }
    return 0;
}

接下来是状压dp

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;

char s[15];
int a[15];

int main()
{
    int n;
    scanf("%d",&n);
    memset(a,inf,sizeof(a));
    for(int i = 0;i < n;++i)
    {
        int m;
        scanf("%d %s",&m,s);
        int len = strlen(s);
        int ans = 0;
        for(int j = 0;j < len;++j)
        {
            switch(s[j]){
                case 'A':ans |= 1;break;
                case 'B':ans |= 2;break;
                case 'C':ans |= 4;break;
            }
        }
        a[ans] = min(a[ans],m);
        for(int j = 1;j <= 7;++j)
        {
            if(a[j] + m < a[(ans | j)]){
                a[(ans | j)] = a[j] + m;
            }
        }
    }
    if(a[7] == inf){
        printf("-1\n");
    }else{
        printf("%d\n",a[7]);
    }
    return 0;
}