NOIP2000提高组复赛C 单词接龙

题目链接：https://ac.nowcoder.com/acm/contest/248/C

题目大意：

　　略

分析：

　　注意点：1.前缀和后缀的公共部分应该选最短的。2.如果两个字符串前缀和后缀的公共部分恰好是其中一个字符串，那么这两个字符串不能合并。

代码如下：

#include <bits/stdc++.h>
using namespace std;
 
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
 
#define pii pair<int,int>
#define piii pair<pair<int,int>,int>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
 
inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
     
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
}
 
inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}
 
typedef long long LL;
const int maxN = 1e5 + 7;
 
int n, ans;
string str[21];
vector< pii > nexts[21];
int cnt[21];
string st;
 
// Common prefix and suffix
int CPS(int x, int y) {
    int ret = 0, i = str[x].size()-1, j = 1;
     
    while(i >= 0 && j < str[y].size()){
        if(str[x].substr(i, str[x].size()) == str[y].substr(0, j)) {
            ret = j;
            break;
        }
        --i;
        ++j;
    }
    if(ret == str[x].size() || ret == str[y].size()) ret = 0;
    return ret;
}
 
void dfs(int x, int ret) {
    bool flag = true;
    rep(i, nexts[x].size()) {
        int nt = nexts[x][i].se;
        if(cnt[nt] >= 2) continue;
        flag = false;
         
        ++cnt[nt];
         
        dfs(nt, ret + str[nt].size() - nexts[x][i].fi);
         
        --cnt[nt];
    }
    if(flag) ans = max(ans, ret);
}
 
int main(){
    scanf("%d", &n);
    For(i, 1, n) cin >> str[i];
    cin >> st;
    For(i, 1, n) {
        For(j, 1, n) {
            int t = CPS(i, j);
            if(t) nexts[i].push_back(mp(t, j));
        }
    }
    For(j, 1, n) {
        if(st[0] == str[j][0])nexts[0].push_back(mp(0, j));
    }
    dfs(0, 0);
     
    printf("%d
", ans);
    return 0;
}