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 << "]" << "\n";
     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-;
 const int inf = 1e9 + ;
 const LL mod = 1e9 + ;
 const int maxN = 1e5 + ;
 const LL ONE = ;

 struct Queen{
     int r, c, cnt = ;
 }; 

 // 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[];

 // 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 = ;
     For(i, , m) {
         if(fcnt(Q[i], Q[i - ])) {
             if(cnt == ) Q[i - ].cnt += ;
             else Q[i - ].cnt += ;
             ++cnt;
         }
         else {
             if(cnt > )Q[i - ].cnt += ;
             cnt = ;
         }
     }
     if(cnt > ) Q[m].cnt += ;
     //For(i, 1, m) cout << Q[i].cnt << " ";
     //cout << endl;
 }

 int main(){
     INIT();
     cin >> n >> m;
     For(i, , m) cin >> Q[i].r >> Q[i].c;

     // 处理左右
     sort(Q + , Q + m + , cmp1);
     CountDir(cnt1);
     // 处理上下
     sort(Q + , Q + m + , cmp2);
     CountDir(cnt2);
     // 处理左上，左下
     sort(Q + , Q + m + , cmp3);
     CountDir(cnt3);
     // 处理右上，右下
     sort(Q + , Q + m + , cmp4);
     CountDir(cnt4);

     For(i, , m) ++ans[Q[i].cnt];
     Rep(i, ) cout << ans[i] << " ";
     cout << endl;
     return ;
 }