方法：dfs 状态压缩
方法比较明显，就是一个基本的回溯问题。据说直接做会超时，然而我还是过了。。
code:
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <string>
6 #include <vector>
7 #include <stack>
8 #include <bitset>
9 #include <cstdlib>
10 #include <cmath>
11 #include <set>
12 #include <list>
13 #include <deque>
14 #include <map>
15 #include <queue>
16 #include <fstream>
17 #include <cassert>
18 #include <unordered_map>
19 #include <unordered_set>
20 #include <cmath>
21 #include <sstream>
22 #include <time.h>
23 #include <complex>
24 #include <iomanip>
25 #define Max(a,b) ((a)>(b)?(a):(b))
26 #define Min(a,b) ((a)<(b)?(a):(b))
27 #define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
28 #define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
29 #define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
30 #define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
31 #define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
32 #define FOREACH(a,b) for (auto &(a) : (b))
33 #define rep(i,n) FOR(i,0,n)
34 #define repn(i,n) FORN(i,1,n)
35 #define drep(i,n) DFOR(i,n-1,0)
36 #define drepn(i,n) DFOR(i,n,1)
37 #define MAX(a,b) a = Max(a,b)
38 #define MIN(a,b) a = Min(a,b)
39 #define SQR(x) ((LL)(x) * (x))
40 #define Reset(a,b) memset(a,b,sizeof(a))
41 #define fi first
42 #define se second
43 #define mp make_pair
44 #define pb push_back
45 #define all(v) v.begin(),v.end()
46 #define ALLA(arr,sz) arr,arr+sz
47 #define SIZE(v) (int)v.size()
48 #define SORT(v) sort(all(v))
49 #define REVERSE(v) reverse(ALL(v))
50 #define SORTA(arr,sz) sort(ALLA(arr,sz))
51 #define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
52 #define PERMUTE next_permutation
53 #define TC(t) while(t--)
54 #define forever for(;;)
55 #define PINF 1000000000000
56 #define newline ‘\n‘
57
58 #define test if(1)if(0)cerr
59 using namespace std;
60 using namespace std;
61 typedef vector<int> vi;
62 typedef vector<vi> vvi;
63 typedef pair<int,int> ii;
64 typedef pair<double,double> dd;
65 typedef pair<char,char> cc;
66 typedef vector<ii> vii;
67 typedef long long ll;
68 typedef unsigned long long ull;
69 typedef pair<ll, ll> l4;
70 const double pi = acos(-1.0);
71
72
73 string board[15];
74 int n;
75 bool c[15] = {false}, d[2][31]={false};
76 ll ans = 0;
77 void dfs(int pos)
78 {
79 if (pos == n)
80 {
81 ++ans;
82 return;
83 }
84 for (int i = 0; i < n; ++i)
85 {
86 if (c[i] || d[0][n-1+pos-i] || d[1][pos+i] || board[pos][i] == ‘*‘) continue;
87 c[i] = d[0][n-1+pos-i] = d[1][pos+i] = true;
88 dfs(pos+1);
89 c[i] = d[0][n-1+pos-i] = d[1][pos+i] = false;
90 }
91 }
92 int main()
93 {
94 ios::sync_with_stdio(false);
95 int kase = 0;
96 while (cin >> n && n)
97 {
98 rep(i, n) cin >> board[i];
99 ans = 0;
100 dfs(0);
101 cout << "Case " << ++kase << ": ";
102 cout << ans << newline;
103 }
104 }
看到别人的题解，使用状态压缩来做的，学到了。
code：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <string>
6 #include <vector>
7 #include <stack>
8 #include <bitset>
9 #include <cstdlib>
10 #include <cmath>
11 #include <set>
12 #include <list>
13 #include <deque>
14 #include <map>
15 #include <queue>
16 #include <fstream>
17 #include <cassert>
18 #include <unordered_map>
19 #include <unordered_set>
20 #include <cmath>
21 #include <sstream>
22 #include <time.h>
23 #include <complex>
24 #include <iomanip>
25 #define Max(a,b) ((a)>(b)?(a):(b))
26 #define Min(a,b) ((a)<(b)?(a):(b))
27 #define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
28 #define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
29 #define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
30 #define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
31 #define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
32 #define FOREACH(a,b) for (auto &(a) : (b))
33 #define rep(i,n) FOR(i,0,n)
34 #define repn(i,n) FORN(i,1,n)
35 #define drep(i,n) DFOR(i,n-1,0)
36 #define drepn(i,n) DFOR(i,n,1)
37 #define MAX(a,b) a = Max(a,b)
38 #define MIN(a,b) a = Min(a,b)
39 #define SQR(x) ((LL)(x) * (x))
40 #define Reset(a,b) memset(a,b,sizeof(a))
41 #define fi first
42 #define se second
43 #define mp make_pair
44 #define pb push_back
45 #define all(v) v.begin(),v.end()
46 #define ALLA(arr,sz) arr,arr+sz
47 #define SIZE(v) (int)v.size()
48 #define SORT(v) sort(all(v))
49 #define REVERSE(v) reverse(ALL(v))
50 #define SORTA(arr,sz) sort(ALLA(arr,sz))
51 #define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
52 #define PERMUTE next_permutation
53 #define TC(t) while(t--)
54 #define forever for(;;)
55 #define PINF 1000000000000
56 #define newline ‘\n‘
57
58 #define test if(1)if(0)cerr
59 using namespace std;
60 using namespace std;
61 typedef vector<int> vi;
62 typedef vector<vi> vvi;
63 typedef pair<int,int> ii;
64 typedef pair<double,double> dd;
65 typedef pair<char,char> cc;
66 typedef vector<ii> vii;
67 typedef long long ll;
68 typedef unsigned long long ull;
69 typedef pair<ll, ll> l4;
70 const double pi = acos(-1.0);
71
72
73 int board[15];
74 int n;
75 ll ans = 0;
76 int ALL;
77 void dfs(int pos, int c, int d1, int d2)
78 {
79 if (pos == n)
80 {
81 ++ans;
82 return;
83 }
84 int nxt = ~(board[pos] | c | d1 | d2);
85 int lb = nxt & -nxt & ALL;
86 while (lb)
87 {
88 dfs(pos+1, c|lb, (d1|lb)<<1, (d2|lb)>>1);
89 nxt -= lb;
90 lb = nxt & -nxt & ALL;
91 }
92 }
93
94 int main()
95 {
96 ios::sync_with_stdio(false);
97 cin.tie(0);
98 int kase = 0;
99 while (cin >> n && n)
100 {
101 ALL = (1<<n)-1;
102 rep(i, n)
103 {
104 board[i] = 0;
105 string str; cin >> str;
106 rep(j, n) if (str[j] == ‘*‘) board[i] |= (1<<j);
107 }
108 ans = 0;
109 dfs(0, 0, 0, 0);
110 cout << "Case " << ++kase << ": ";
111 cout << ans << newline;
112 }
113 }
时间： 2024-12-22 17:01:24