UVa 11795 Mega Man's Mission (状压DP)

题意：你最初只有一个武器，你需要按照一定的顺序消灭n个机器人(n<=16)。每消灭一个机器人将会得到他的武器。

每个武器只能杀死特定的机器人。问可以消灭所有机器人的顺序方案总数。

析：dp[s] 表示已经杀死 s 这个状态的机器人有多少种方案，然后挨着枚举每个机器人，在枚举机器人要保证能够杀死该机器人。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = (1<<16) + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL dp[maxn];
int p[maxn], a[20];
char s[20];

int main(){
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d", &n);
        scanf("%s", s);
        int st = 0;
        for(int i = 0; i < n; ++i)  if(s[i] == '1')  st |= (1<<i);
        for(int i = 0; i < n; ++i){
            scanf("%s", s);
            a[i] = 0;
            for(int j = 0; j < n; ++j)  if(s[j] == '1')  a[i] |= (1<<j);
        }

        int all = (1<<n) - 1;
        for(int i = 0; i <= all; ++i){
            p[i] = st;
            for(int j = 0; j < n; ++j)  if(i & (1<<j))  p[i] |= a[j];
        }
        memset(dp, 0, sizeof dp);
        dp[0] = 1;
        for(int i = 1; i <= all; ++i)
            for(int j = 0; j < n; ++j)
                if((p[i^(1<<j)]&i)&(1<<j))  dp[i] += dp[i^(1<<j)];
        printf("Case %d: %lld\n", kase, dp[all]);
    }
    return 0;
}