CF1043D Mysterious Crime

思路：

参考了http://codeforces.com/blog/entry/62797，把第一个序列重标号成1，2，3，...，n，在剩下的序列中寻找形如x, x + 1, x + 2, ...的连续子段即可。

实现：

 #include <iostream>
 #include <vector>
 #include <cstring>
 using namespace std;
 typedef long long ll;

 const int INF = 0x3f3f3f3f;

 int a[][], reach[][], n, m;

 int main()
 {
     while (scanf("%d %d", &n, &m) != EOF)
     {
         memset(reach, , sizeof reach);
         for (int i = ; i <= m; i++)
         {
             for (int j = ; j <= n; j++)
                 scanf("%d", &a[i][j]);
         }
         vector<int> v(n + );
         for (int i = ; i <= n; i++) v[a[][i]] = i;
         for (int i = ; i <= m; i++)
         {
             for (int j = ; j <= n; j++)
                 a[i][j] = v[a[i][j]];
         }
         for (int i = ; i <= n; i++) reach[][i] = n;
         for (int i = ; i <= m; i++)
         {
             int j = , start = ;
             while (j <= n)
             {
                 while (j +  <= n && a[i][j + ] == a[i][j] + ) j++;
                 while (start <= j) { reach[i][a[i][start]] = a[i][j]; start++; }
                 start = ++j;
             }
         }
         ll ans = ;
         for (int i = ; i <= n; i++)
         {
             int minn = INF;
             for (int j = ; j <= m; j++)
                 if (reach[j][i])
                     minn = min(minn, reach[j][i] - i + );
             if (minn != INF) ans += minn;
         }
         printf("%lld\n", ans);
     }
     return ;
 }