题干

C++实现——复杂度极高的深度优先遍历(容易超时)

#include <iostream>
#include <vector>
using namespace std;//sa+sb = sum
//sb-sa = diffint sum = 0;
int diff = 0;void DFSFindMinDiff(vector<int> &arr,int pos,int sa) { //sa表示a集合的元素和if(pos == arr.size()) {return;}//arr[pos]不放入a集合DFSFindMinDiff(arr,pos+1,sa);//arr[pos]放入a集合int newdiff;//记录当前差值if(2*(sa+arr[pos]) - sum > 0) {//sa>sbnewdiff = 2*(sa+arr[pos]) - sum;} else {//sa<sbnewdiff = sum - 2*(sa+arr[pos]);}if(newdiff < diff) {diff = newdiff;}DFSFindMinDiff(arr,pos+1,sa+arr[pos]);
}int main() {vector<int> arr;int i;//EOF对应键盘的ctrl+cwhile(scanf("%d",&i) != EOF) {arr.push_back(i);}for(int i = 0; i<arr.size(); ++i) {sum+=arr[i];}diff = sum;//一开始，a集合一个元素也没有，全都在b集合中DFSFindMinDiff(arr,0,0);int sa = (sum-diff)/2;int sb = sa+diff;//sa更小printf("%d %d\n",sb,sa);return 0;
}

C++实现——优化策略(剪枝)

• arr排序，从大到小
• 先执行加入arr[pos]，再执行不加入的
• 一开始sa<sb，某个时刻sa>sb，就可以终止了
• 当diff=1时，可以提前终止
• 当arr[pos] > sum/2时，可以终止

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;int sum = 0;//sa+sb = sum
int diff = 0;//sb-sa = diff
bool exitFlag = false;//用来记录是否提前退出
bool compare(int lhs,int rhs){return lhs>rhs;
}void DFSFindMinDiff(vector<int> &arr,int pos,int sa) { //sa表示a集合的元素和if(pos == arr.size() || exitFlag == true) {return;}//arr[pos]放入a集合int newdiff;//记录当前差值if(2*(sa+arr[pos]) - sum > 0) {//sa>sbnewdiff = 2*(sa+arr[pos]) - sum;} else {//sa<sbnewdiff = sum - 2*(sa+arr[pos]);}if(newdiff < diff) {diff = newdiff;if(diff == 0||diff==1||2 * arr[pos] > sum){//这三种情况，可提前终止exitFlag = true;}}if(2*(sa+arr[pos]) - sum < 0){DFSFindMinDiff(arr,pos+1,sa+arr[pos]);}//arr[pos]不放入a集合DFSFindMinDiff(arr,pos+1,sa);
}int main() {vector<int> arr;int i;//EOF对应键盘的ctrl+cwhile(scanf("%d",&i) != EOF) {arr.push_back(i);}for(int i = 0; i<arr.size(); ++i) {sum+=arr[i];}diff = sum;//一开始，a集合一个元素也没有，全都在b集合中sort(arr.begin(),arr.end(),compare);//先从大到小排序DFSFindMinDiff(arr,0,0);int sa = (sum-diff)/2;int sb = sa+diff;//sa更小printf("%d %d\n",sb,sa);return 0;
}