子集和问题

来源：互联网发布：nodejs 连接数据库编辑：程序博客网时间：2024/05/22 01:40
input.txt
5 10
2 4 6 5 2
#include <iostream>#include <fstream>using namespace std;class subset{public:subset(int* best, int* arr, int n, int result);void backtrack();   bool Place(int k);   int* arr;//序列各个个数的值   int n;//数字总个数   int *x;//当前解   int *best;//最优解   int r;//剩下的序列和   int nresult;//当前序列的总和   int result;//目的序列和};subset::subset(int* best, int* arr, int n, int result){this->best = best;this->arr = arr;this->n = n;this->result = result;x = new int[n + 1];r = 0;for (int i = 0; i < n; i++){r += arr[i];}}bool subset::Place(int k){//第k个不装的情况下当前序列和剩下的和是否能组成resultif ((nresult + r) < result)return false;return true;}void subset::backtrack(){//向左孩子出生条件nresult+arr[k]<result,右孩子出生条件Place()x[1] = 0;int k = 1;nresult = 0;//搜索子树while (true){while ((k <= n) && (nresult + arr[k] <= result)){//从k节点开始一直向左r -= arr[k];nresult += arr[k];x[k] = 1;k++;}if (k > n){//到达终点if (nresult == result){//找到节点结束for (int j = 1; j <= n; j++){best[j] = x[j];}return;}else{//返回到最后后一次进入队列的节点k--;//退回到叶子节点r += arr[k];//释放资源nresult -= arr[k];//释放资源k--;//找最后一次加入的节点while (k > 0 && !x[k]){//从右子树返回r += arr[k];k--;}nresult -= arr[k];x[k] = 0;k++;}}else{//进入右子树r -= arr[k];x[k] = 0;k++;}}}int main(){//数据读取ifstream in("input.txt", ios::in);if (!in.is_open()){cout << "Error opening file";exit(1);}char ch;int n = in.get() - '0';//子集个数while ((ch = in.get()) != ' '){n = n * 10 + ch - '0';}int* arr = new int[n+1];int result = in.get() - '0';//记录序列和while ((ch = in.get()) != '\n'){result = result * 10 + ch - '0';}int i = 1;while (!in.eof()){arr[i] = in.get() - '0';while( ((ch = in.get()) != ' ')&&(ch!= '\n')&&(ch!=-1)){arr[i] = arr[i] * 10 + ch - '0';}i++;}/*for (int i = 0; i < n; i++){cout << arr[i] << "  " <<  endl;}*/in.close();//数据处理int* best = new int[n+1]();//标记最优解subset demo(best,arr,n,result);demo.backtrack();for (int i = 1; i <= n; i++){if (best[i] == 1)cout << arr[i] << endl;}return 0;}
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