[codeforces 200 A Cinema]暴力，优化

题意大致是这样的：有一个有n行、每行m个格子的矩形，每次往指定格子里填石子，如果指定格子里已经填过了，则找到与其曼哈顿距离最小的格子，然后填进去，有多个的时候依次按x、y从小到大排序然后取最小的。输出每次填的格子的坐标。

思路：这道题出自Codeforces Round #126 (Div. 2)是个暴力优化的题。如果指定格子未填，则填到里面。否则枚举曼哈顿距离，然后枚举格子找答案。裸的暴力太慢了，主要是因为每次曼哈顿距离都是从1开始搜索，如果每次指定的坐标都是同一个，则做了大量的重复工作。不妨用一个数组r[x][y]表示与(x,y)这个格子曼哈顿距离不超过r[x][y]的格子全部被填过了。r数组满足这样一个关系，r[x][y]>=r[x'][y']-dist{(x,y),(x',y')}，每次使用r[x][y]之前，用(x,y)周围的一些点更新下就行了。复杂度比较模糊，必须承认，这种优化简直太神，对于极端数据和随机数据都灰常之快

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/* ******************************************************************************** */ #include <iostream>                                                                 // #include <cstdio>                                                                   // #include <cmath>                                                                    // #include <cstdlib>                                                                  // #include <cstring>                                                                  // #include <vector>                                                                   // #include <ctime>                                                                    // #include <deque>                                                                    // #include <queue>                                                                    // #include <algorithm>                                                                // #include <map>                                                                      // #include <cmath>                                                                    // using namespace std;                                                                //                                                                                     // #define pb push_back                                                                // #define mp make_pair                                                                // #define X first                                                                     // #define Y second                                                                    // #define all(a) (a).begin(), (a).end()                                               // #define fillchar(a, x) memset(a, x, sizeof(a))                                      //                                                                                     // typedef pair<int, int> pii;                                                         // typedef long long ll;                                                               // typedef unsigned long long ull;                                                     //                                                                                     // #ifndef ONLINE_JUDGE                                                                // void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}    // void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>                    // void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;          // while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>      // void print(const T t){cout<<t<<endl;}template<typename F,typename...R>              // void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>   // void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}   // #endif // ONLINE_JUDGE                                                              // template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}        // template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}        // template<typename T>                                                                // void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}            // template<typename T>                                                                // void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}            //                                                                                     // const double PI = acos(-1.0);                                                       // const int INF = 1e9 + 7;                                                            //                                                                                     // /* -------------------------------------------------------------------------------- */ const int maxn = 2e3 + 7;   bool plot[maxn][maxn]; int r[maxn][maxn]; int n, m, k;   int check(int row, int col) {     if (col >= 0 && col < m) return true;     return false; }   int main() { #ifndef ONLINE_JUDGE     freopen("in.txt", "r", stdin);     //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE       cin >> n >> m >> k;     for (int i = 0; i < k; i ++) {         int x, y;         scanf("%d%d", &x, &y);         x --; y --;         if (!plot[x][y]) {             printf("%d %d\n", x + 1, y + 1);             plot[x][y] = true;             continue;         }         for (int d = 1; d <= 3; d ++) {             int Max = min(n, x + d + 1);             for (int row = max(0, x - d); row < Max; row ++) {                 int t = d - abs(x - row), col1 = y - t, col2 = y + t;                 if (check(row, col1)) umax(r[x][y], r[row][col1] - d);                 if (check(row, col2)) umax(r[x][y], r[row][col2] - d);             }         }         for (int d = r[x][y] + 1; ; d ++) {             int Max = min(n, x + d + 1);             bool ok = false;             for (int row = max(0, x - d); row < Max; row ++) {                 int t = d - abs(x - row), col1 = y - t, col2 = y + t;                 if (check(row, col1) && !plot[row][col1]) {                     ok = true;                     printf("%d %d\n", row + 1, col1 + 1);                     plot[row][col1] = true;                     break;                 }                 if (check(row, col2) && !plot[row][col2]) {                     ok = true;                     printf("%d %d\n", row + 1, col2 + 1);                     plot[row][col2] = true;                     break;                 }             }             if (ok) {                 r[x][y] = d - 1;                 break;             }         }     }     return 0; } /* ******************************************************************************** */