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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 | /* ******************************************************************************** */#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
", 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
", row + 1, col1 + 1); plot[row][col1] = true; break; } if (check(row, col2) && !plot[row][col2]) { ok = true; printf("%d %d
", row + 1, col2 + 1); plot[row][col2] = true; break; } } if (ok) { r[x][y] = d - 1; break; } } } return 0;}/* ******************************************************************************** */ |