POJ3977 Subset 折半枚舉

標籤：algorithm   poj   演算法   

    題目大意是給定N個數的集合，從這個集合中找到一個非空子集，使得該子集元素和的絕對值最小，如果有多個答案，輸出元素個數最少的那個。   

N最多為35，如果直接枚舉顯然是不行的。但是如果我們將這些數分成兩半後再枚舉的話，最多有2^18(262144)，此時我們兩半枚舉後的結果進行排序後再二分搜尋一下就可以了。複雜度為O(nlogn) n最多2^18。

#include <stdio.h>#include <vector>#include <math.h>#include <string.h>#include <string>#include <iostream>#include <queue>#include <list>#include <algorithm>#include <stack>#include <map>using namespace std;struct MyStruct{long long res;int i;};int compp(const void* a1, const void* a2){long long dif = ((MyStruct*)a1)->res - ((MyStruct*)a2)->res;if (dif > 0){return 1;}else if (dif == 0){return 0;}elsereturn -1;}MyStruct res[2][300000];inline long long absll(long long X){if (X < 0){return X * (-1);}elsereturn X;}int main(){int n;#ifdef _DEBUGfreopen("d:\\in.txt", "r", stdin);#endiflong long values[36];while (scanf("%d", &n) != EOF){if (n == 0){break;}for (int i = 0; i < n; i++){scanf("%I64d", &values[i]);}int maxn = n - n / 2;int maxm = n - maxn;memset(res, 0, sizeof(res));for (int i = 0; i < 1 << maxn; i++){res[0][i].i = i;for (int k = 0; k < 19; k++){if ((i >> k) & 1){res[0][i].res += values[k];}}}qsort(res[0], 1 << maxn, sizeof(MyStruct), compp);for (int i = 0; i < 1 << maxm; i++){res[1][i].i = i;for (int k = 0; k < 19; k++){if ((i >> k) & 1){res[1][i].res += values[k + maxn];}}}qsort(res[1], 1 << maxm, sizeof(MyStruct), compp);long long minvalue = 1000000000000000LL;int mink = 32;int l = 0;int r = (1 << maxm);for (int i = 0; i < 1 << maxn; i++){l = 0;int curk = 0;for (int k = 0; k < maxn; k++){if ((res[0][i].i >> k) & 1){curk++;}}while (r - l > 1){int mid = (l + r) / 2;long long sum = res[1][mid].res + res[0][i].res;if (sum > 0){r = mid;}elsel = mid;}l = l >= 1 ? l - 1 : l;for (int k = l; k < (1 << maxm);k++){int curm = 0;for (int m = 0; m < maxm; m++){if ((res[1][k].i >> m) & 1){curm++;}}if (curm == 0 && curk == 0){continue;}long long sum = res[1][k].res + res[0][i].res;if (absll(sum) < minvalue){mink = curm + curk;minvalue = absll(sum);}else if (absll(sum) == minvalue){mink = min(mink, curk + curm);}else if (sum > 0){break;}}}printf("%I64d %d\n", minvalue, mink);}return 0;}

POJ3977 Subset 折半枚舉