UVA 10817

题目：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=20&page=show_problem&problem=1758

状态压缩的DP，dp[i][st]表示状态为st考虑后面i个人所有人最小花费，

因为每个科目有三种状态，可以用一个三进制数表示，

状态不是很多，所以可以把预先把每个数的三进制预处理出来，

决策为选和不选。

#include<bits/stdc++.h>
using namespace std;
const int maxs = 8;
const int MAXSTA = 6561+5;
const int maxn = 101;
int base[maxs+1],trp[MAXSTA][maxs+1];
int dp[maxn][MAXSTA];
int tch[maxn],cost[maxn],sa;
int s,m,n;
const int INF = 0x3fffffff;
void preDeal()
{
    base[0] = 1;
    for(int i = 1; i <= maxs; i++){
        base[i] = 3*base[i-1];
    }
    for(int i = 1; i < MAXSTA; i++){
        int x = i, cnt = 0;
        while(x){
            trp[i][cnt++] = x%3;
            x /= 3;
        }
    }
}

int dfs(int i,int st)
{
    if(i == n) return st == sa? 0:INF;
    int &ans = dp[i][st];
    if(~ans) return ans;
    ans = dfs(i+1,st);
    //int nst = st;
    for(int j = 0; j < s; j++){
        if(trp[tch[i]][j] && trp[st][j] != 2){
            st += base[j];
        }
    }
    ans = min(ans,dfs(i+1,st)+cost[i]);
    return ans;
}

int work()
{
    sa = base[s]-1;
    int st = 0,sum = 0;
    for(int i = 0; i < m; i++){
        int c; scanf("%d",&c);
        sum += c;
        while(getchar()!='
'){
            int sub = getchar()-'1';
            int t = trp[st][sub] ;
            if(t != 2){
                st += base[sub];
            }
        }
    }
    for(int i = 0; i < n; i++){
        scanf("%d",cost+i);
        tch[i] = 0;
        while(getchar()!='
'){
            tch[i] += base[getchar()-'1'];
        }
    }
    memset(dp,-1,sizeof(dp));
    return sum + dfs(0,st);
}

int main()
{
    //freopen("in.txt","r",stdin);
    preDeal();
    while(scanf("%d%d%d",&s,&m,&n),s){
        printf("%d
",work());
    }
    return 0;
}