[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;
}
/* ******************************************************************************** */
posted @ 2015-08-04 06:11  jklongint  阅读(300)  评论(0)    收藏  举报