102 - Ecological Bin Packing

最新推荐文章于 2021-03-17 17:18:42 发布

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本文介绍了一种通过枚举所有可能的垃圾箱排列来解决生态垃圾箱打包问题的方法，并实现了一个C++程序来找到使垃圾箱移动次数最少且字典序最小的方案。

此题本身不难，因为对于输入，只有6种可能的组合。直接用暴力法枚举所有可能就可以了。

主要的问题在于保证输出所有最优解中字典序最少的一个。

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// 102_Ecological_Bin_Packing.cpp : Defines the entry point for the console application. // #include <set> #include <map> #include <list> #include <queue> #include <stack> #include <vector> #include <string> #include <sstream> #include <cstdio> #include <cmath> #include <limits> #include <utility> #ifndef ONLINE_JUDGE #include <fstream> std::ifstream cin("in.txt"); std::ofstream cout("out.txt"); // std::ofstream cout(stdout); #else #include <iostream> #endif typedef long long llong; typedef unsigned long long ullong; typedef long double ldouble; using namespace std; #define BROWN 0 #define CLEAR 2 #define GREEN 1 #define SUM 3 vector<vector<int> > recycleBins(3); int CountMoves(int c1, int c2, int c3) { return recycleBins[0][SUM] - recycleBins[0][c1] + recycleBins[1][SUM] - recycleBins[1][c2] + recycleBins[2][SUM] - recycleBins[2][c3]; } int moveType[6][3] = { { BROWN, CLEAR, GREEN }, { BROWN, GREEN, CLEAR }, { CLEAR, BROWN, GREEN }, { CLEAR, GREEN, BROWN }, { GREEN, BROWN, CLEAR }, { GREEN, CLEAR, BROWN }, }; char colorNames[] = "BGC"; int main(int argc, char* argv[]) { bool bFinished = false; recycleBins[0].resize(4); recycleBins[1].resize(4); recycleBins[2].resize(4); for(;;) { for(int i = 0; i < 3 && !bFinished; ++i) { recycleBins[i][SUM] = 0; for(int j = 0; j < 3; ++j) { if(!(cin>> recycleBins[i][j])) { bFinished = true; break; } recycleBins[i][SUM] += recycleBins[i][j]; } } if(bFinished) break; int minMove = numeric_limits<int>::max(); int minIndex = -1; for(int i = 0; i < 6; ++i) { int moves = CountMoves(moveType[i][0], moveType[i][1], moveType[i][2]); if(moves < minMove) { minMove = moves; minIndex = i; } } for(int i = 0; i < 3; ++i) cout<< colorNames[ moveType[minIndex][i] ]; cout<< " "<< minMove<< "/n"; } return 0; }

102 - Ecological Bin Packing