題目鏈接：https://ac.nowcoder.com/acm/contest/248/C
題目大意：
略
分析：
注意點：1.前綴和后綴的公共部分應該選最短的。2.如果兩個字符串前綴和后綴的公共部分恰好是其中一個字符串，那么這兩個字符串不能合并。
代碼如下：
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define rep(i,n) for (int i = 0; i < (n); ++i) 5 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 6 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 7 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 8 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 9 10 #define pr(x) cout << #x << " = " << x << " " 11 #define prln(x) cout << #x << " = " << x << endl 12 13 #define ALL(x) x.begin(),x.end() 14 #define INS(x) inserter(x,x.begin()) 15 16 #define ms0(a) memset(a,0,sizeof(a)) 17 #define msI(a) memset(a,inf,sizeof(a)) 18 19 #define pii pair<int,int> 20 #define piii pair<pair<int,int>,int> 21 #define mp make_pair 22 #define pb push_back 23 #define fi first 24 #define se second 25 26 inline int gc(){ 27 static const int BUF = 1e7; 28 static char buf[BUF], *bg = buf + BUF, *ed = bg; 29 30 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 31 return *bg++; 32 } 33 34 inline int ri(){ 35 int x = 0, f = 1, c = gc(); 36 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 37 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 38 return x*f; 39 } 40 41 typedef long long LL; 42 const int maxN = 1e5 + 7; 43 44 int n, ans; 45 string str[21]; 46 vector< pii > nexts[21]; 47 int cnt[21]; 48 string st; 49 50 // Common prefix and suffix 51 int CPS(int x, int y) { 52 int ret = 0, i = str[x].size()-1, j = 1; 53 54 while(i >= 0 && j < str[y].size()){ 55 if(str[x].substr(i, str[x].size()) == str[y].substr(0, j)) { 56 ret = j; 57 break; 58 } 59 --i; 60 ++j; 61 } 62 if(ret == str[x].size() || ret == str[y].size()) ret = 0; 63 return ret; 64 } 65 66 void dfs(int x, int ret) { 67 bool flag = true; 68 rep(i, nexts[x].size()) { 69 int nt = nexts[x][i].se; 70 if(cnt[nt] >= 2) continue; 71 flag = false; 72 73 ++cnt[nt]; 74 75 dfs(nt, ret + str[nt].size() - nexts[x][i].fi); 76 77 --cnt[nt]; 78 } 79 if(flag) ans = max(ans, ret); 80 } 81 82 int main(){ 83 scanf("%d", &n); 84 For(i, 1, n) cin >> str[i]; 85 cin >> st; 86 For(i, 1, n) { 87 For(j, 1, n) { 88 int t = CPS(i, j); 89 if(t) nexts[i].push_back(mp(t, j)); 90 } 91 } 92 For(j, 1, n) { 93 if(st[0] == str[j][0])nexts[0].push_back(mp(0, j)); 94 } 95 dfs(0, 0); 96 97 printf("%d\n", ans); 98 return 0; 99 }
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