UVa 1399 Puzzle
方法:AC自动机
先把禁止的string插入ac自动机中,然后再这个自动机上求最长的合法路径。如果出现环或者最长路径长度为0,则输出“No";否则输出最长路径。
code:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 const int maxnode = 4e5+5, sigma_size = 26; 73 int n, K; 74 struct AC 75 { 76 int sz, g[maxnode][sigma_size]; 77 int fail[maxnode], last[maxnode]; 78 void init() 79 { 80 sz = 1; Reset(g[0], 0); 81 } 82 int idx(char c) 83 { 84 return c-'A'; 85 } 86 int newnode() 87 { 88 last[sz] = 0; Reset(g[sz], 0); 89 return sz++; 90 } 91 void insert(const string &str) 92 { 93 int n = str.length(), cur = 0; 94 rep(i, n) 95 { 96 int c = idx(str[i]); 97 if (!g[cur][c]) g[cur][c] = newnode(); 98 cur = g[cur][c]; 99 } 100 last[cur] = 1; 101 } 102 void get_fail() 103 { 104 queue<int> q; 105 fail[0] = 0; 106 rep(i, sigma_size) 107 { 108 int u = g[0][i]; 109 if (u) 110 { 111 fail[u] = 0; 112 q.push(u); 113 } 114 } 115 while (!q.empty()) 116 { 117 int cur = q.front(); q.pop(); 118 rep(i, K) 119 { 120 int u = g[cur][i]; 121 if (!u) 122 { 123 g[cur][i] = g[fail[cur]][i]; continue; 124 } 125 q.push(u); int nxt = fail[cur]; 126 while (nxt && !g[nxt][i]) nxt = fail[nxt]; 127 fail[u] = g[nxt][i]; 128 last[u] |= last[fail[u]]; 129 } 130 } 131 } 132 } solver; 133 bitset<maxnode> vis; 134 int dp[maxnode], jump[maxnode]; 135 136 int dfs(int cur) 137 { 138 if (vis[cur]) return -1; 139 int &ans = dp[cur]; 140 if (ans == -1) 141 { 142 vis[cur] = true; 143 ans = 0; 144 drep(i, K) 145 { 146 int nxt = solver.g[cur][i]; 147 if (!solver.last[nxt]) 148 { 149 int tmp = dfs(nxt); 150 if (tmp == -1) return -1; 151 if (tmp+1 > ans) 152 { 153 ans = tmp+1; 154 jump[cur] = i; 155 } 156 } 157 } 158 } 159 vis[cur] = 0; 160 return ans; 161 } 162 163 int main() 164 { 165 ios::sync_with_stdio(false); 166 cin.tie(0); 167 int T; cin >> T; 168 repn(kase, T) 169 { 170 cin >> K >> n; 171 solver.init(); 172 Reset(dp, -1); 173 Reset(jump, -1); 174 vis.reset(); 175 string line; 176 rep(i, n) 177 { 178 cin >> line; 179 solver.insert(line); 180 } 181 solver.get_fail(); 182 int ans = dfs(0); 183 if (ans == -1 || !ans) 184 { 185 cout << "No\n"; 186 } 187 else 188 { 189 190 int cur = 0; 191 for(;;) 192 { 193 if (jump[cur] == -1) break; 194 cout << (char)(jump[cur]+'A'); 195 cur = solver.g[cur][jump[cur]]; 196 } 197 cout << newline; 198 } 199 } 200 } 201 202 /* 203 204 3 2 4 AAA AB BA BB 2 4 AAA BBB ABAB BBAA 3 7 AA ABA BAC BB BC CA CC 205 206 */