ACM模版

描述

题解

先将指针排序，然后获取指针刻度差值，接着，找到一个起点存储起来，用作后续的比较。这里起点的查找标准具体看函数，不做解释。求得有多少段相同的序列，累加相同状态的钟表对儿数即可（代码One）。

这里可以进行一个比较好的优化，使用哈希+map（代码Two）。

代码

One：

#include <iostream>
#include <algorithm>
#include <cstdio>

using namespace std;

const int MAXN = 5e3 + 5;
const int INF = 0x3f3f3f3f;

int N, M, P;

int hands[MAXN];
int hand[MAXN];

struct clo
{
    int hands[MAXN];
} Clock[MAXN];

bool cmp(clo a, clo b)
{
    for (int i = 0; i < M; i++)
    {
        if (a.hands[i] != b.hands[i])
        {
            return a.hands[i] < b.hands[i];
        }
    }
    return false;
}

// 检测是否相等
bool cmp_(clo a, clo b)
{
    for (int i = 0; i < M; i++)
    {
        if (a.hands[i] != b.hands[i])
        {
            return false;
        }
    }
    return true;
}

void squ(int key, int *hand)
{
    int MIN = INF;
    int pos = 0;
    for (int i = 0; i < M; i++)
    {
        if (hand[i] < MIN)
        {
            MIN = hand[i];
            pos = i;
        }
    }
    for (int i = pos + 1; i < M; i++)
    {
        if (hand[i] == hand[pos])
        {
            for (int j = 1; j < M; j++)
            {
                if (hand[(i + j) % M] > hand[(pos + j) % M])
                {
                    break;
                }
                else if (hand[(i + j) % M] < hand[(pos + j) % M])
                {
                    pos = i;
                }
            }
        }
    }
    int tag = 0;
    for (int i = pos; i < M; i++)
    {
        Clock[key].hands[tag++] = hand[i];
    }
    for (int i = 0; i < pos; i++)
    {
        Clock[key].hands[tag++] = hand[i];
    }
    return ;
}

int main(int argc, const char * argv[])
{
    cin >> N >> M >> P;
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            scanf("%d", hands + j);
        }
        sort(hands, hands + M);
        hand[0] = P - hands[M - 1] + hands[0];
        for (int j = 1; j < M; j++)
        {
            hand[j] = hands[j] - hands[j - 1];
        }
// sort(Clock[i].hands, Clock[i].hands + M); // WA，需要按特殊规则排
        squ(i, hand);
    }
    sort(Clock, Clock + N, cmp);

    int pos = 0;
    int ans = 0;
    for (int i = 1; i <= N; i++)
    {
        if (!cmp_(Clock[pos], Clock[i]))
        {
            int num = i - pos;
            ans += num * (num - 1) / 2;
            pos = i;
        }
    }
    std::cout << ans << '\n';
    return 0;
}

Two：

#include <iostream>
#include <algorithm>
#include <map>
#include <cstdio>

using namespace std;

const int N = 500 + 10;

int n, m, p;
int A[N][N], B[N][N];

int minID[N];

void input()
{
    for (int i = 0; i < n; ++i)
    {
        for (int j = 0; j < m; ++j)
        {
            scanf("%d", &A[i][j]);
        }
    }
    for (int i = 0; i < n; ++i)
    {
        sort(A[i], A[i] + m);
        // 求指针差
        for (int j = 0; j < m; ++j)
        {
            B[i][j] = (A[i][(j + 1) % m] - A[i][j] + p ) % p;
        }
    }
}

// 寻找起点
int getMin(int *C)
{
    int i = 0, j = 1, k = 0;
    while (i < m && j < m && k < m)
    {
        int t = C[(i + k) % m] - C[(j + k) % m];
        if (!t)
        {
            k++;
        }
        else
        {
            if (t > 0)
            {
                i += k + 1;
            }
            else
            {
                j += k + 1;
            }
            if (i == j)
            {
                j++;
            }
            k = 0;
        }
    }
    return i < j ? i : j;
}

map<int, int> S;
int SEED = 133331;

void solve()
{
    for (int i = 0; i < n; ++i)
    {
        minID[i] = getMin(B[i]);
    }
    S.clear();
    int ans = 0;
    for (int i = 0; i < n; ++i)
    {
        int tmp = 1;
        for (int k = 0; k < m; ++k)
        {
            tmp = tmp * SEED + B[i][(minID[i] + k) % m];    // 哈希
        }
        ans += S[tmp];
        S[tmp]++;
    }
    cout << ans << endl;
    return ;
}

int main()
{
    while (~scanf("%d%d%d", &n, &m, &p))
    {
        input();
        solve();
    }
    return 0;
}