CCF NOI1055. 走廊 (C++)

1055. 走廊

题目描述

有一个M行N列的教室座位中，有D对同学总爱凑在一起讲话。现老师要用走廊隔开他们。但只能在行之间加入K条走廊，在列中加入L条走廊，问加在哪里能使效果最佳。（一对爱讲话的同学只有左右相邻或上下相邻）。

输入

第一行，有5各用空格隔开的整数，分别是M，N，K，L，D（2<=N，M<=1000，0<=K<M，0<=L<N，D<=2000）。接下来D行，每行有4个用空格隔开的整数，第i行的4个整数Xi，Yi，Pi，Qi，表示坐在位置(Xi,Yi)与(Pi,Qi)的两个同学会交头接耳（输入保证他们前后相邻或者左右相邻）。输入数据保证最优方案的唯一性。

输出

共两行。第一行包含K个整数，a1a2……aK，表示第a1行和a1+1行之间、第a2行和第a2+1行之间、…、第aK行和第aK+1行之间要开辟通道，其中ai< ai+1，每两个整数之间用空格隔开（行尾没有空格）。第二行包含L个整数，b1b2……bk，表示第b1列和b1+1列之间、第b2列和第b2+1列之间、…、第bL列和第bL+1列之间要开辟通道，其中bi< bi+1，每两个整数之间用空格隔开（行尾没有空格）。若有多组答案，输出字典序最小的一组。

样例输入

4 5 1 2 3
4 2 4 3
2 3 3 3
2 5 2 4

样例输出

2
2 4

数据范围限制

C++代码

#include <iostream>
#include <cstring>
#include <vector>
#include <map>
#include <iterator>
#include <algorithm>
#include <cassert>using namespace std;//#define CPP_STL_MAP_VECTOR class WP
{public:int weight; // weightint roc;    // row or column
};const int MAX_M = 1000;
const int MAX_N = 1000;
const int MAX_D = 2000;
const int MAX_L = MAX_N-1; // L < N
const int MAX_K = MAX_M-1; // K < M#ifdef CPP_STL_MAP_VECTORvector<WP> all_row_corridors_vec;
vector<WP> all_col_corridors_vec;map<int, int> all_row_corridors_mp;
map<int, int> all_col_corridors_mp;#elseWP all_row_corridors[MAX_L];
WP all_col_corridors[MAX_K];#endifbool comp(const WP &wp1, const WP &wp2)
{if (wp1.weight == wp2.weight){return wp1.roc < wp2.roc;  // sort ascending by roc        }else{return wp1.weight > wp2.weight;  // sort descending by weight}
}int main()
{int M, N, K, L, D;cin >> M >> N >> K >> L >> D;assert(N>=2 && N <= MAX_N && N > L);assert(M>=2 && M <= MAX_M);assert(K>=0 && K < M);assert(L>=0 && L < N);assert(D>=0 && D <= MAX_D && D > L);#ifndef CPP_STL_MAP_VECTOR for(int row=1; row<=MAX_L; row++){all_row_corridors[row-1].roc    = row;all_row_corridors[row-1].weight = 0;}for(int col=1; col<=MAX_K; col++){all_col_corridors[col-1].roc    = col;all_col_corridors[col-1].weight = 0;}
#endifint Xi, Yi, Pi, Qi;int max_col = 0, max_row = 0;for (int row=1; row<=D; row++){cin >> Xi >> Yi >> Pi >> Qi;// check each pair students, and// calculate the corresponding corridors' weight if (Xi == Pi){#ifdef CPP_STL_MAP_VECTOR all_col_corridors_mp[min(Yi, Qi)]++;
#elseall_col_corridors[min(Yi, Qi) - 1].weight++;max_col = max(max_col, min(Yi, Qi));
#endif}else if (Yi == Qi){#ifdef CPP_STL_MAP_VECTORall_row_corridors_mp[min(Xi, Pi)]++;
#elseall_row_corridors[min(Xi, Pi) - 1].weight++;max_row = max(max_row, min(Xi, Pi));
#endif}}#ifdef CPP_STL_MAP_VECTORWP wp;map<int, int>::iterator curr;for (curr=all_col_corridors_mp.begin();curr!=all_col_corridors_mp.end();curr++){wp.roc    = curr->first;wp.weight = curr->second;all_col_corridors_vec.push_back(wp);}for (curr=all_row_corridors_mp.begin();curr!=all_row_corridors_mp.end();curr++){wp.roc    = curr->first;wp.weight = curr->second;all_row_corridors_vec.push_back(wp);      }sort(all_row_corridors_vec.begin(), all_row_corridors_vec.end(), comp);sort(all_col_corridors_vec.begin(), all_col_corridors_vec.end(), comp);
#elsesort(all_row_corridors, all_row_corridors+max_row, comp);sort(all_col_corridors, all_col_corridors+max_col, comp);
#endif// sort and output corridor(s) between rowsvector<int> output_corridors_between_rows;for(int row=1;row<=K; row++){#ifdef CPP_STL_MAP_VECTORoutput_corridors_between_rows.push_back(all_row_corridors_vec[row-1].roc);
#elseoutput_corridors_between_rows.push_back(all_row_corridors[row-1].roc);
#endif}sort(output_corridors_between_rows.begin(), output_corridors_between_rows.end());copy(output_corridors_between_rows.begin(), output_corridors_between_rows.end(), ostream_iterator<int>(cout, " "));cout << endl;// sort and output corridor(s) between columnsvector<int> output_corridors_between_cols;for(int col=1;col<=L; col++){#ifdef CPP_STL_MAP_VECTORoutput_corridors_between_cols.push_back(all_col_corridors_vec[col-1].roc);
#elseoutput_corridors_between_cols.push_back(all_col_corridors[col-1].roc);
#endif}sort(output_corridors_between_cols.begin(),output_corridors_between_cols.end());copy(output_corridors_between_cols.begin(),output_corridors_between_cols.end(), ostream_iterator<int>(cout, " "));cout << endl;return 0;
}