标签:style blog io ar color os sp for on
11.7 有个马戏团正在设计叠罗汉的表演节目,一个人要站在另一人的肩膀上。处于实际和美观的考虑,在上面的人要比下面的人矮一点、轻一点。已知马戏团每个人的高度和重量,请编写代码计算叠罗汉最多能叠几个人。
如果要保持相对顺序不变,那么不能直接排序。
C++实现代码:
#include<iostream> #include<vector> using namespace std; struct people { int weight; int lenght; people(int w,int l):weight(w),lenght(l){} }; bool isValid(vector<people> &path,people &p) { if(path.empty()) return true; people tmp=path.back(); return p.lenght<tmp.lenght&&p.weight<tmp.weight; } void helper(vector<people> &pp,int start,int &longSequence,vector<people> &path) { if(start==pp.size()) { if(longSequence<path.size()) longSequence=path.size(); return; } int i; for(i=start;i<pp.size();i++) { //该条件可以不成立,因此不能进入递归 if(isValid(path,pp[i])) { path.push_back(pp[i]); helper(pp,i+1,longSequence,path); path.pop_back(); } } } int getIncreasingSequence(vector<people> &pp) { int longSequence=0; vector<people> path; helper(pp,0,longSequence,path); return longSequence; } int main() { vector<people> p={people(1,1),people(2,2),people(3,3),people(4,4)}; cout<<getIncreasingSequence(p)<<endl; }
标签:style blog io ar color os sp for on
原文地址:http://www.cnblogs.com/wuchanming/p/4154092.html