【题解】UVA10817：Headmaster‘s Headache

@vjudge
记录有0、1、2个老师教的课的状态$s0,s1,s2$，并记录当前枚举到第几个老师$t$
$dfs(t,s0,s1,s2)$
用记忆化，状态方面，$s0,s1,s2$中保存两个就好了，剩下一个用另外两个算出来，所以状态合理

Code：

#include <bits/stdc++.h>
#define maxn 310
#define maxm 210
using namespace std;
int kase, power[25], s, n, m, vis[maxm][maxn][maxn], dp[maxm][maxn][maxn], w[maxm], v[maxm];

int dfs(int t, int s0, int s1, int s2){
	if (t > n + m) return s2 == power[s] - 1 ? 0 : 1e9;
	if (vis[t][s1][s2] == kase) return dp[t][s1][s2];
	vis[t][s1][s2] = kase;
	int sum = 1e9;
	if (t > m) sum = min(sum, dfs(t + 1, s0, s1, s2));
	int x = s0 & v[t], y = s1 & v[t];
	sum = min(sum, dfs(t + 1, s0 ^ x, s1 ^ y | x, s2 | y) + w[t]);
	return dp[t][s1][s2] = sum;
}

int main(){
	power[0] = 1;
	for (int i = 1; i <= 10; ++i) power[i] = power[i - 1] << 1;
	while (1){
		scanf("%d%d%d", &s, &m, &n);
		if (s == 0 && n == 0 && m == 0) break;
		++kase;
		for (int i = 1; i <= n + m; ++i){
			scanf("%d", &w[i]);
			v[i] = 0;
			while (1){
				char c = getchar();
				if (isdigit(c)) v[i] |= power[(c ^ 48) - 1];
				if (c == '\n') break;
			}
		}
		printf("%d\n", dfs(1, power[s] - 1, 0, 0));
	}
}

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