UVA-11795 Mega Man's Mission (状压DP)

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思路：

• 求出每个状态i(已杀死机器人集合)能杀死的机器人数kill[i]，即已死机器人的武器加上初始武器又能杀死的机器人的集合；
• 每个状态i都只能往包含在kill[i]中，且不包含在s中的机器人j扩展，即i -> i|(1 << j)；
• dp[i]代表到达该状态的顺序总数，i等于所有前一个合法状态的和，易知dp[0] = 1，且只有从0状态扩展的状态才是合法的状态；

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Code:

 

#include<cstdio>
#include<cstring>
using namespace std;
#define M(a) memset(a, 0, sizeof(a))
#define INF 0x3f3f3f3f
const int maxn = (1 << 16) + 10;
int s[maxn], kill[maxn];
long long dp[maxn];

int main() {

    int T, n, kase = 0;
    scanf("%d", &T);
    while(T--) {
        char str[20];
        M(dp), M(kill), M(s);
        scanf("%d%s", &n, str);
        for (int i = 0; i < n; ++i)
            if (str[i] == '1')
                kill[0] += (1 << i);
        for (int i = 0; i < n; ++i) {
            scanf("%s", str);
            for (int j = 0; j < n; ++j)
                if (str[j] == '1') 
                    s[i] |= (1 << j);
        }
        int all = (1 << n) - 1;
        for (int i = 0; i <= all; ++i) {
            kill[i] = kill[0];
            for (int j = 0; j < n; ++j)
                if (i & (1 << j))
                    kill[i] |= s[j];
        }
        dp[0] = 1;
        for (int i = 0; i < all; ++i)
            for (int j = 0; j < n; ++j)
                if (((kill[i] & (1 << j))) && !(i & (1 << j)))
                    dp[i|(1 << j)] += dp[i];
        printf("Case %d: %lld\n", ++kase, dp[all]);
    }

    return 0;
}