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[11][100001], reach[11][100001], n, m;

int main()
{
    while (scanf("%d %d", &n, &m) != EOF)
    {
        memset(reach, 0, sizeof reach);
        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
                scanf("%d", &a[i][j]);
        }
        vector<int> v(n + 1);
        for (int i = 1; i <= n; i++) v[a[1][i]] = i;
        for (int i = 2; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
                a[i][j] = v[a[i][j]];
        }
        for (int i = 1; i <= n; i++) reach[1][i] = n;
        for (int i = 2; i <= m; i++)
        {
            int j = 1, start = 1;
            while (j <= n)
            {
                while (j + 1 <= n && a[i][j + 1] == a[i][j] + 1) j++;
                while (start <= j) { reach[i][a[i][start]] = a[i][j]; start++; }
                start = ++j;
            }
        }
        ll ans = 0;
        for (int i = 1; i <= n; i++)
        {
            int minn = INF;
            for (int j = 1; j <= m; j++)
                if (reach[j][i])
                    minn = min(minn, reach[j][i] - i + 1);
            if (minn != INF) ans += minn;
        }
        printf("%lld
", ans);
    }
    return 0;
}