51Nod－1282－时钟

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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;}

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