USACO算法系列十五——shoping

    题目：http://www.nocow.cn/index.php/Translate:USACO/shopping

    题目很长，大致意思是买n种商品，然后这几种商品按照不同的组合可以有不同的折扣，要我们按照最优的折扣来买东西。看到这道题很自然想到的是背包问题，但是这道题包含的信息明显复杂的多。让我想起了一句话“数据结构是算法的基础，好的数据结构能够很大程度上减少算法的复杂度。”

    我使用的第一种方法是，深度搜索剪枝方法，代码如下：

#include <iostream>#include <fstream>#define TYPE 1000#define SIZE 100using namespace std;struct Good{int id; int num;};struct GoodPrivilige{int num;Good g[5];int value;};//记录商品价格int value[TYPE] = {0};GoodPrivilige p[SIZE];int pl;Good need[5];int nl;int result = 0x7FFFFFFF;void calResult(int level, int money){if (level >= pl){for (int i=0;i <nl; i ++){money += need[i].num * value[need[i].id];}if (money < result){result = money;}}else{int n = 0;int m = money;while (true){int flag = true;for (int i=0; i < nl; i ++){if (need[i].num < 0){flag = false;break;}}//检验是否有效//检查有效性 并进行剪枝if(flag && m < result ){calResult(level+1,m);}elsebreak;n++;//使用方案的数量for (int i=0; i < p[level].num; i ++){int id = p[level].g[i].id;for (int j=0; j < nl; j ++){if (id == need[j].id){need[j].num -= p[level].g[i].num;break;}//end if}//end for j}//end for im += p[level].value;}//end while//还原使用方案的数量for (int i=0; i < p[level].num; i ++){int id = p[level].g[i].id;for (int j=0; j < nl; j ++){if (id == need[j].id){need[j].num += n * p[level].g[i].num;break;}//end if}//end for j}//end for i}//end else}int main(){ifstream fin("shopping.in");ofstream fout("shopping.out"); //输入数据//输入方案fin >> pl;for (int i =0; i< pl; i ++){fin >> p[i].num;for (int j=0; j < p[i].num; j ++){fin >> p[i].g[j].id;fin >> p[i].g[j].num;}fin >> p[i].value;}//输入要购买的物品fin >> nl;for (int i=0; i < nl; i ++){fin >> need[i].id >> need[i].num;fin >> value[need[i].id];}calResult(0,0);fout << result << endl;return 0;}

    很可惜的是在Test10的时候，就超时了。

Executing... Test 1: TEST OK [0.000 secs, 3024 KB] Test 2: TEST OK [0.011 secs, 3024 KB] Test 3: TEST OK [0.011 secs, 3024 KB] Test 4: TEST OK [0.000 secs, 3024 KB] Test 5: TEST OK [0.000 secs, 3024 KB] Test 6: TEST OK [0.022 secs, 3024 KB] Test 7: TEST OK [0.022 secs, 3024 KB] Test 8: TEST OK [0.011 secs, 3024 KB] Test 9: TEST OK [0.011 secs, 3024 KB] > Run 10: Execution error: Your program (`shopping') used more than the allotted runtime of 1 seconds (it ended or was stopped at 1.674 seconds) when presented with test case 10. It used 3024 KB of memory.

    没办法，只好对其进行优化，跟背包的改进算法是一样的，只是用到了一个5维的数组，用于记录迭代前进的过程。代码如下：

#include <iostream>#include <fstream>#define TYPE 1000#define SIZE 100using namespace std;struct Good{int id; int num;};struct GoodPrivilige{int num;Good g[5];int value;};//记录商品价格int value[TYPE] = {0};GoodPrivilige p[SIZE];int pl;Good need[5];int nl;int result;int table[6][6][6][6][6] = {0};int getPrivilige(int index,int id){for (int i=0; i < p[index].num; i ++){if (id == p[index].g[i].id){return p[index].g[i].num;}}return 0;}void calResult(){for (int n = 0; n < pl; n ++){int p1 = getPrivilige(n, need[0].id);int p2 = getPrivilige(n, need[1].id);int p3 = getPrivilige(n, need[2].id);int p4 = getPrivilige(n,need[3].id);int p5 = getPrivilige(n,need[4].id);for (int i=p1; i <= need[0].num; i ++){for (int j=p2; j <= need[1].num; j ++){for (int k=p3; k <= need[2].num; k ++){for (int l =p4; l <= need[3].num; l ++){for (int m = p5; m <= need[4].num; m ++){if(table[i][j][k][l][m] > table[i-p1][j-p2][k-p3][l-p4][m-p5] + p[n].value){table[i][j][k][l][m] = table[i-p1][j-p2][k-p3][l-p4][m-p5] + p[n].value;}}//end m}//end l}//end k}//end j}//end i}//end n;}int main(){ifstream fin("shopping.in");ofstream fout("shopping.out");//输入数据//输入方案fin >> pl;for (int i =0; i< pl; i ++){fin >> p[i].num;for (int j=0; j < p[i].num; j ++){fin >> p[i].g[j].id;fin >> p[i].g[j].num;}fin >> p[i].value;}//输入要购买的物品fin >> nl;for (int i=0; i < nl; i ++){fin >> need[i].id >> need[i].num;fin >> value[need[i].id];}//初始化for (int i=0; i <= need[0].num; i ++){for (int j=0; j <= need[1].num; j ++){for (int k=0; k <= need[2].num; k ++){for (int l =0; l <= need[3].num; l ++){for (int m = 0; m <= need[4].num; m ++){table[i][j][k][l][m] = i * value[need[0].id] + j * value[need[1].id] + k * value[need[2].id] + l * value[need[3].id] + m * value[need[4].id];}//end m}//end l}//end k}//end j}//end icalResult();result = table[need[0].num][need[1].num][need[2].num][need[3].num][need[4].num];fout << result << endl;return 0;}

    速度还不错。结果如下：

Executing... Test 1: TEST OK [0.022 secs, 3052 KB] Test 2: TEST OK [0.000 secs, 3052 KB] Test 3: TEST OK [0.000 secs, 3052 KB] Test 4: TEST OK [0.000 secs, 3052 KB] Test 5: TEST OK [0.011 secs, 3052 KB] Test 6: TEST OK [0.000 secs, 3052 KB] Test 7: TEST OK [0.000 secs, 3052 KB] Test 8: TEST OK [0.000 secs, 3052 KB] Test 9: TEST OK [0.011 secs, 3052 KB] Test 10: TEST OK [0.000 secs, 3052 KB] Test 11: TEST OK [0.011 secs, 3052 KB] Test 12: TEST OK [0.000 secs, 3052 KB]All tests OK.