题意：给一个长度为16的字符串，每次从里面删掉一个回文序列，求最少需要几次才能删掉所有字符
思路：二进制表示每个字符的状态，那么从1个状态到另一个状态有两种转移方式，一是枚举所有合法的回文子序列，判断是否是当前状态的子状态，再转移，二是枚举当前状态的所有子状态来转移。前者最坏复杂度O(2^16*2^16) = O(几十亿)，而后者最坏只有(i:1->16)Σ2iC(16,i) = O(几千万）。
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#include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; #ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} #endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} template<typename T> void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} template<typename T> void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} const double PI = acos(-1.0); const int INF = 1e9 + 7; /* -------------------------------------------------------------------------------- */ char s[20]; int n, dp[1 << 20]; void init() { for (int i = 0; i < (1 << n); i ++) dp[i] = INF; dp[0] = 1; for (int i = 1; i < (1 << n); i ++) { int j, k; for (j = n - 1; j >= 0; j --) { if (i & (1 << j)) break; } for (k = 0; k < n; k ++) { if (i & (1 << k)) break; } if (s[j] == s[k] && dp[i ^ (1 << j) ^ (1 << k)] < INF) dp[i] = dp[i ^ (1 << j) ^ (1 << k)]; if (j == k) dp[i] = 1; } } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int T; cin >> T; while (T --) { scanf("%s", s); n = strlen(s); init(); for (int i = 1; i < (1 << n); i ++) { for (int j = i; j; j = (j - 1) & i) { umin(dp[i], dp[i ^ j] + dp[j]); } } cout << dp[(1 << n) - 1] << endl; } return 0; } |