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UVa 1401 Remember the word
方法:dp + trie
比较明显可以想出一个dp,对于给定的string str,d[i] = 表示str.substr(i) 的方法数, d[str.length()] = 1, d[i] = sum(d[i+x.length()], x 是 str.substr(i) 的prefix),最后答案是d[0]。状态转移的时候,可以通过走由可行x组成的trie来判断。
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 const int maxnode = 400000+5; 73 const int sigma_size = 26; 74 75 76 struct Trie 77 { 78 int g[maxnode][sigma_size]; 79 int val[maxnode]; 80 int sz; 81 int newnode() 82 { 83 Reset(g[sz], 0); 84 val[sz] = 0; 85 return sz++; 86 } 87 void init() { sz = 1; Reset(g[0], 0); }; 88 int idx(char c) { return c-‘a‘; }; 89 90 void insert(const string &str, int v) 91 { 92 int u = 0, n = str.length(); 93 for (int i = 0; i < n; ++i) 94 { 95 int c = idx(str[i]); 96 if (!g[u][c]) g[u][c] = newnode(); 97 u = g[u][c]; 98 } 99 val[u] = v; 100 } 101 }; 102 103 Trie solver; 104 string str; 105 string input[4000]; 106 int n; 107 const int mod = 20071027; 108 int d[300001]; 109 int main() 110 { 111 ios::sync_with_stdio(false); 112 cin.tie(0); 113 int kase = 0; 114 while (cin >> str >> n) 115 { 116 solver.init(); 117 for (int i = 0; i < n; ++i) cin >> input[i]; 118 for (int i = 0; i < n; ++i) solver.insert(input[i], 1); 119 int len = str.length(); 120 Reset(d, 0); 121 d[len] = 1; 122 for (int i = len-1; i >= 0; --i) 123 { 124 int u = 0; 125 for (int j = i; j < len; ++j) 126 { 127 int c = solver.idx(str[j]); 128 if (!solver.g[u][c]) 129 { 130 break; 131 } 132 u = solver.g[u][c]; 133 if (solver.val[u]) d[i] = (d[i] + d[j+1]) % mod; 134 } 135 } 136 cout << "Case " << ++kase << ": " << d[0] << newline; 137 } 138 }
UVa 1401 Remember the word
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