PKU_campus_2018_D Chocolate

思路：

题目链接http://poj.openjudge.cn/practice/C18D/

kruskal过程中使用乘法原理计数。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int MAXN = , INF = 0x3f3f3f3f;
 const ll MOD = 1e9 + ;
 string s[MAXN];
 int d[MAXN][MAXN];
 int par[MAXN];
 ll num[MAXN];
 struct edge
 {
     int a, b, l;
 };
 edge es[MAXN * MAXN];
 void init(int n)
 {
     for (int i = ; i < n; i++) { par[i] = i; num[i] = ; }
 }
 int find(int x)
 {
     if (par[x] == x) return x;
     return par[x] = find(par[x]);
 }
 int uni(int x, int y)
 {
     x = find(x); y = find(y);
     if (x == y) return x;
     par[x] = y; return y;
 }
 int edit_dis(string a, string b)
 {
     int la = a.length(), lb = b.length();
     for (int i = ; i <= la; i++) d[i][] = i;
     for (int i = ; i <= lb; i++) d[][i] = i;
     for (int i = ; i <= la; i++)
     {
         for (int j = ; j <= lb; j++)
         {
             if (a[i - ] == b[j - ]) d[i][j] = d[i - ][j - ];
             else d[i][j] = min(d[i][j - ], d[i - ][j]) + ;
         }
     }
     return d[la][lb];
 }
 bool cmp(edge & x, edge & y)
 {
     return x.l < y.l;
 }
 bool check(int x, int cnt)
 {
     int minn = INF, maxn = -INF;
     for (int i = ; i < cnt; i++)
     {
         int px = find(es[i].a), py = find(es[i].b);
         if (px == x && py == x) maxn = max(maxn, es[i].l);
         else if ((px == x && py != x) || (px != x && py == x))
             minn = min(minn, es[i].l);
     }
     return maxn < minn;
 }
 int main()
 {
     int t, n;
     cin >> t;
     while (t--)
     {
         cin >> n;
         init(n);
         for (int i = ; i < n; i++) cin >> s[i];
         int cnt = ;
         for (int i = ; i < n; i++)
         {
             for (int j = i + ; j < n; j++)
             {
                 es[cnt].a = i; es[cnt].b = j; es[cnt].l = edit_dis(s[i], s[j]);
                 cnt++;
             }
         }
         sort(es, es + cnt, cmp);
         ll ans = ; int tot = n;
         for (int i = ; i < cnt; i++)
         {
             int px = find(es[i].a), py = find(es[i].b);
             if (px == py) continue;
             int tmp = uni(es[i].a, es[i].b);
             num[tmp] = num[px] * num[py] % MOD;
             if (check(tmp, cnt)) num[tmp] = (num[tmp] + ) % MOD;
             tot--;
             if (tot == ) ans = num[tmp];
         }
         cout << ans << endl;
     }
     return ;
 }