1053. Path of Equal Weight (30)

Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. Theweight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L.

Now given any weighted tree, you are supposed to find all the paths with their weights equal to a given number. For example, let's consider the tree showed in Figure 1: for each node, the upper number is the node ID which is a two-digit number, and the lower
number is the weight of that node. Suppose that the given number is 24, then there exists 4 different paths which have the same given weight: {10 5 2 7}, {10 4 10}, {10 3 3 6 2} and {10 3 3 6 2}, which correspond to the red edges in Figure 1.



Figure 1
Input Specification:

Each input file contains one test case. Each case starts with a line containing 0 < N <= 100, the number of nodes in a tree, M (< N), the number of non-leaf nodes, and 0 < S < 230, the given weight number. The next line contains N positive numbers
where Wi (<1000) corresponds to the tree node Ti. Then M lines follow, each in the format:

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 00.

Output Specification:

For each test case, print all the paths with weight S in non-increasing order. Each path occupies a line with printed weights from the root to the leaf in order. All the numbers must be separated by a space with no extra space at the end
of the line.

Note: sequence {A1, A2, ..., An} is said to begreater than sequence {B1, B2, ..., Bm} if there exists 1 <= k < min{n, m} such that Ai = Bi for i=1, ...
k, and Ak+1 > Bk+1.

Sample Input:

20 9 24
10 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 2
00 4 01 02 03 04
02 1 05
04 2 06 07
03 3 11 12 13
06 1 09
07 2 08 10
16 1 15
13 3 14 16 17
17 2 18 19

Sample Output:

10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2

按要求输入树，然后检索根节点到叶子结点路径长等于指定长度的路径，按一定排序，输出沿途途径。

调用的元素时，需要初始化， 并且->first 代表map<first,second>中的first，->second 代表其中的second。这是map容器和其他容器不同的地方。

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今天又看了下，发现排序算法有点蛋疼，直接从前到后，从大倒下排序输出即可，而且也不用 专门建一个容器存放vector的，因为vector本身就是容器。

#include <iostream>
#include <string>
#include <vector>
#include <map>
#include<algorithm>
using namespace std;

class tree
{
public:
int weight;
int num;
tree *father;
bool haschild;
tree(int w,int n)
{
haschild=false;
father=NULL;
weight=w;
num=n;
}
tree()
{

haschild=false;
father=NULL;
}
};
struct out
{
vector<int> t;
};

bool compare(out a,out b);
int main()
{
map<int,tree> trees;
int N,M,S;
cin>>N>>M>>S;
for(int i=0;i<N;i++)//存入
{
int w;
cin>>w;
tree t=tree(w,i);
trees[i]=t;
}
for(int i=0;i<M;i++)//找个各自的父亲节点
{
int num,k;
cin>>num>>k;
map<int,tree>::iterator tfather=trees.find(num);
tfather->second.haschild=true;
for(int j=0;j<k;j++)
{
int child;
cin>>child;
map<int,tree>::iterator tchild=trees.find(child);
tchild->second.father=&tfather->second;
}
}
vector <out> rightout;
for(int i=0;i<N;i++)
{
int allweight=0;
map<int,tree>::iterator thistree=trees.find(i);//找到节点
if(thistree->second.haschild==true)//只对叶子结点检索
continue;
out count;//存入输出结果
tree fathertree=thistree->second;
count.t.push_back(fathertree.weight);
allweight+=fathertree.weight;
//计算路经长，为了方便排序，这里用了三个容器，一个是记录路径的vector，一个是存有vector的结点，一个是存有结点的vector，略乱
while(fathertree.father!=NULL)
{
fathertree=*fathertree.father;
count.t.push_back(fathertree.weight);
allweight+=fathertree.weight;
}
if(S==allweight)
{
rightout.push_back(count);
}
}
sort(rightout.begin(),rightout.end(),compare);
for(int i=0;i<rightout.size();i++)
{
int j;
for(j=0;j<rightout[i].t.size()-1;j++)
cout<<rightout[i].t[rightout[i].t.size()-j-1]<<" ";//这里要当心不要多输出一个空格，为了这个调了5分钟= =
cout<<rightout[i].t[rightout[i].t.size()-j-1]<<endl;
}
system("pause");
return 0;
}
bool compare(out a,out b)//按要求排序
{
int i=0;
while(a.t[a.t.size()-i-1]==b.t[b.t.size()-i-1]&&i<a.t.size()-1&&i<b.t.size()-1)//这里要注意长度限制，不然很容易越界！
i++;
return a.t[a.t.size()-i-1]>b.t[b.t.size()-i-1];
}