牛客练习赛13F m皇后

题目链接：https://ac.nowcoder.com/acm/contest/70/F

题目大意：

　　略

分析：

　　可以分成四步计算冲突：水平方向，垂直方向，左斜线方向，右斜线方向。只要会处理水平方向，其余同理。

代码如下：

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
 
#define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl

#define SORT(c, s, t) sort(c + s, c + t + 1)
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))
 
#define pii pair<int,int>
#define piii pair<pair<int,int>,int>
#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef set< int > SI;
typedef vector< int > VI;
const double EPS = 1e-10;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;
 
struct Queen{
    int r, c, cnt = 0;
}; 

// 先按行排序，再按列排序 
inline bool cmp1(const Queen &x, const Queen &y) {
    if(x.r == y.r) return x.c < y.c;
    return x.r < y.r;
}

// 先按列排序，再按行排序 
inline bool cmp2(const Queen &x, const Queen &y) {
    if(x.c == y.c) return x.r < y.r;
    return x.c < y.c;
}

// 先按r - c排序，再按r + c排序 
inline bool cmp3(const Queen &x, const Queen &y) {
    if(x.r - x.c == y.r - y.c) return x.r + x.c < y.r + y.c;
    return x.r - x.c < y.r - y.c;
}

// 先按r + c排序，再按r - c排序 
inline bool cmp4(const Queen &x, const Queen &y) {
    if(x.r + x.c == y.r + y.c) return x.r - x.c < y.r - y.c;
    return x.r + x.c < y.r + y.c;
}

// 行方向计数辅助函数
inline bool cnt1(const Queen &x, const Queen &y) {
    return x.r == y.r;
}

// 列方向计数辅助函数
inline bool cnt2(const Queen &x, const Queen &y) {
    return x.c == y.c;
}

// 左斜线方向计数辅助函数
inline bool cnt3(const Queen &x, const Queen &y) {
    return x.r - x.c == y.r - y.c;
}

// 右斜线方向计数辅助函数
inline bool cnt4(const Queen &x, const Queen &y) {
    return x.r + x.c == y.r + y.c;
}

int n, m;
Queen Q[maxN];
int ans[9];

// 方向计数函数 
void CountDir(bool (*fcnt)(const Queen&,const Queen&)) {
    //cout << endl;
    //For(i, 1, m) cout << MP(Q[i].r, Q[i].c);
    int cnt = 1;
    For(i, 2, m) {
        if(fcnt(Q[i], Q[i - 1])) {
            if(cnt == 1) Q[i - 1].cnt += 1;
            else Q[i - 1].cnt += 2;
            ++cnt;
        }
        else {
            if(cnt > 1)Q[i - 1].cnt += 1;
            cnt = 1;
        } 
    }
    if(cnt > 1) Q[m].cnt += 1;
    //For(i, 1, m) cout << Q[i].cnt << " ";
    //cout << endl;
}

int main(){
    INIT();
    cin >> n >> m;
    For(i, 1, m) cin >> Q[i].r >> Q[i].c;
    
    // 处理左右 
    sort(Q + 1, Q + m + 1, cmp1);
    CountDir(cnt1);
    // 处理上下 
    sort(Q + 1, Q + m + 1, cmp2);
    CountDir(cnt2);
    // 处理左上，左下 
    sort(Q + 1, Q + m + 1, cmp3);
    CountDir(cnt3);
    // 处理右上，右下 
    sort(Q + 1, Q + m + 1, cmp4);
    CountDir(cnt4);
    
    For(i, 1, m) ++ans[Q[i].cnt];
    Rep(i, 9) cout << ans[i] << " ";
    cout << endl;
    return 0;
}