poj1014 dfs/背包/类筛法求素数

题意：有分别价值为1,2,3,4,5,6的6种物品，输入6个数字，表示相应价值的物品的数量，问能否将物品分成两份，使两份价值相等，其中每个物品不能切开，只能分给其中的某一方，当输入六个0时，程序结束，总物品的总个数不超过20000。

输出：每个测试用例占三行：
第一行： Collection #k: k为第几组测试用例
第二行：是否能分（具体形式见用例）

第三行：空白（必须注意，否则PE）

算法：

1.Dfs搜索

2.多重背包  ->  0-1背包+二进制拆分

3.类筛法求素数

/*dfs:无回溯WA，无剪枝TLE */#include<iostream>using namespace std;int values[7];bool flag;int halfValue;bool visited[120001];// visited[i]=true：数值i已经计算过，即marbles可以分为i和sum-i，该数组用于剪枝 void init(){for (int i=0; i<=120001; i++){visited[i] = false;} }void output(){cout <<  values[1] << " " <<  values[2] << " " <<  values[3] << " " <<  values[4] << " " <<  values[5] << " " <<  values[6] << " " << endl;} void Dfs(int currVal){if (currVal == halfValue){flag = true;return;}for (int i=6; i>=1; i--){if (values[i] > 0 && !flag){if (currVal + i > halfValue){continue;}else{values[i]--;if (!visited[currVal + i])// 如果 数值currVal + i已经计算过，不要重复计算，如此可以减少大量计算，否则会TLE {Dfs(currVal + i);visited[currVal + i] = true;}values[i]++;// 此处需要回溯，否则0 0 3 0 3 1难过也！ }}}}int main(){int cases = 1;while (cin >> values[1] >> values[2] >> values[3] >> values[4] >> values[5] >> values[6]){int sum = values[1]*1 + values[2]*2 + values[3]*3 + values[4]*4 + values[5]*5 + values[6]*6;if (sum == 0){break;}cout << "Collection #" << cases++ <<':' << endl;if (sum % 2 != 0){            cout << "Can't be divided." << endl << endl;    //注意有空行            continue;}flag = false;halfValue = sum / 2;init();Dfs(0);flag?  cout<<"Can be divided."<<endl<<endl : cout<<"Can't be divided."<<endl<<endl;}return 0;}

/*多重背包 -> 二进制拆分 + 0-1背包 */#include <iostream>#include <string.h>using namespace std;const int MAX=120000;int cnt;// the total number of the marbles after binary splitint marbles[MAX];// marble[i]=j: the value of the ith marble is jint bag[MAX];// bag[i]=j: when the capacity of the bag is i,the max value of the marbles can be put into the bag is jint values[7];// binary split (please refer to《二进制拆分》)void binSplit(int val, int num){int bin_num;int bin_sum = 0;for (int i=0; ; i++){bin_num = 1<<i;if (bin_sum+bin_num > num){break;}marbles[cnt++] = bin_num * val;bin_sum += bin_num ;}int t = val*num - bin_sum*val;if (t > 0){marbles[cnt++] = t; }}// 0-1 bagbool dp(int c){// For each marble which after binary splitfor (int i=0; i<cnt; i++){// For each capacity of j,enumerate every marble i,judge whether it should be put into the bag or not// If put marble i into bag,then bag[j] = bag[j-marbles[i]]+marbles[i]// Surely,in order to ensure marble i can be put into the bag which it's capacity is j,j should greater than marbles[i]// If do not put marble i into bag,then bag[j] is just bag[j]// By comparing the size of bag[j] and bag[j-marbles[i]]+marbles[i],we get the answer......for (int j=c; j>=marbles[i]; j--){if (bag[j-marbles[i]]+marbles[i] > bag[j]){bag[j] =  bag[j-marbles[i]]+marbles[i];}}}return bag[c] == c;}int main(){int cases = 1;int halfValue;while (cin >> values[1] >> values[2] >> values[3] >> values[4] >> values[5] >> values[6]){int sum = values[1]*1 + values[2]*2 + values[3]*3 + values[4]*4 + values[5]*5 + values[6]*6;if (sum == 0){break;}cout << "Collection #" << cases++ <<':' << endl;if (sum % 2 != 0){            cout << "Can't be divided." << endl << endl;    // Output a blank line after each test case            continue;}memset(bag,0,sizeof(bag));memset(marbles,0,sizeof(int));cnt = 0;for (int i=1; i<=6; i++){if (values[i] > 0){binSplit(i,values[i]);} }halfValue = sum / 2;dp(halfValue)? cout<<"Can be divided."<<endl<<endl:cout<<"Can't be divided."<<endl<<endl;}return 0;}

/*类似筛法求素数：设总数为sum，flag[i]=true表示可以分为i和sum-i。显然，flag[0]=true。如果flag[i]=true，而且存在重量为j(1<=j<=6)的marble，那么flag[i+j]=true。而题目要求的是flag[sum/2]。 */#include <iostream>#include <string.h>using namespace std;const int MAX=120000;bool flag[MAX];// flag[i]=ture:marbles可以分为i和sum-i bool tmp_flag[MAX];    int values[7];// values[i]=j：重量为i的marble有j个 int main(){int cases = 1;int halfValue;while (cin >> values[1] >> values[2] >> values[3] >> values[4] >> values[5] >> values[6]){int sum = values[1]*1 + values[2]*2 + values[3]*3 + values[4]*4 + values[5]*5 + values[6]*6;if (sum == 0){break;}cout << "Collection #" << cases++ <<':' << endl;// 如果sum为奇数，肯定不可分 if (sum % 2 != 0){            cout << "Can't be divided." << endl << endl;    //注意有空行            continue;}memset(flag,false,sizeof(flag));memset(tmp_flag,false,sizeof(tmp_flag));flag[0] = true;// for each type of marblehalfValue = sum / 2;for (int i=1; i<=6; i++){if (flag[halfValue]){break;}if (values[i] > 0){ // for each valuefor (int j=0; j<=halfValue; j++){if (tmp_flag[halfValue]){break;}if (flag[j]){for (int k=1; k<=values[i]; k++){if (tmp_flag[halfValue]){break;}if (j+i*k > halfValue){break;}tmp_flag[j+i*k] = true;}}}for (int j=0; j<=halfValue; j++){flag[j] = flag[j] || tmp_flag[j];}}}// outputif (flag[halfValue]){cout<<"Can be divided."<<endl<<endl;}else{cout<<"Can't be divided."<<endl<<endl;}}return 0;}