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A mobile is a type of kinetic sculpture constructed to take advantage of the principle of equilibrium. It consists of a number of rods, from which weighted objects or further rods hang. The objects
hanging from the rods balance each other, so that the rods remain more or less horizontal. Each rod hangs from only one string, which gives it freedom to rotate about the string.
We consider mobiles where each rod is attached to its string exactly in the middle, as in the figure underneath. You are given such a configuration, but the weights on the ends are chosen incorrectly, so that the mobile is not in equilibrium. Since that‘s not aesthetically pleasing, you decide to change some of the weights.
What is the minimum number of weights that you must change in order to bring the mobile to equilibrium? You may substitute any weight by any (possibly non-integer) weight. For the mobile shown in the figure, equilibrium can be reached by changing the middle weight from 7 to 3, so only 1 weight needs to changed.
<expr> ::= <weight> | "[" <expr> "," <expr> "]"with <weight> a positive integer smaller than 109 indicating a weight and [<expr>,<expr>] indicating a rod with the two expressions at the ends of the rod. The total number of rods in the chain from a weight to the top of the mobile will be at most 16.
3 [[3,7],6] 40 [[2,3],[4,5]]
103
本题很有意思，首先要分析发现，通过“平衡”这个关系，一个叶子就可以决定其他所有叶子的大小，进一步思考，也就是说一个节点就决定了整颗二叉树的大小，所以我们就利用这个性质去判断需要修改多少个节点，因为重量不同的节点所算出来的整颗二叉树点总重量必定也是不同的，所以我们只要搜索所有节点，然后去求每个节点所产生的总重，出现次数最多的总重量就代表有最多的叶子节点不需要进行修改。
本题中希望用到这样一种数据结构，能够统计任意类型的数据的出现次数，我们可以采用stl里面的map，遍历和清空的使用方法如下所示,还有一个值得注意的地方就是给定一段字符的s和e，这段字符是一个数字，整理成数字的方法。
#include <iostream>
#include <cstring>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
using namespace std;
map<long long,int>base;
int sum;
string str;
void dfs(int s,int e,int depth)
{
if(str[s]=='[')
{
///cout<<s<<" "<<e<<endl;
int p;
p=0;
for(int i=s+1;i<=e;i++)
{
//cout<<s<<" "<<e<<endl;
if(str[i]=='[') p++;
if(str[i]==']') p--;
if(!p&&str[i]==',')
{
dfs(s+1,i-1,depth+1);
dfs(i+1,e-1,depth+1);
}
}
}
else
{
long long w;
w=0;
for(int i=s;i<=e;i++)
{
w=w*10+str[i]-'0';
}
//cout<<w<<endl;
sum++;
base[w<<depth]++;
//printflag();
}
}
int main()
{
//freopen("input.txt","r",stdin);
int T;
cin>>T;
for(int t=1;t<=T;t++)
{
cin>>str;
int len;
base.clear();
sum=0;
len=str.length();
dfs(0,len-1,0);
int maxn=0;
for(map<long long,int>::iterator it=base.begin();it!=base.end();++it)
{
maxn=max(maxn,it->second);
}
printf("%d\n",sum-maxn);
}
}
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原文地址：http://blog.csdn.net/u013555159/article/details/51356003
