PAT1053

#include<iostream>
#include<vector>
#include<stack>
#include<algorithm>
#include<iterator>
using namespace std;

#define MAX 9999

void record(stack<int> st, vector<vector<int>> &paths)
{
    vector<int> p;
    while(!st.empty())
    {
        p.push_back(st.top());
        st.pop();
    }
    reverse(p.begin(), p.end());
    paths.push_back(p);
}

void DFS_visit(vector<vector<int>> &tree, int S, int start, 
    vector<vector<int>> &paths, stack<int> &st, int sum, vector<int> &weight)
{
    st.push(weight[start]);
    sum += weight[start];
    /*累计权重小于等于S进入判断下一个DFS，
    若大于则不用后续DFS*/
    if(sum <= S)
    {
        /*判断是否是叶子结点*/
        bool flag(false);
        for(int i=0; i<tree.size(); ++i)
            if(tree[start][i] != MAX)
            {
                DFS_visit(tree, S, i, paths, st, sum, weight);
                flag = true;
            }
        if(flag == false)
            if(sum == S)
                record(st, paths);
    }
    sum -= st.top();
    st.pop();
}

void DFS_version(vector<vector<int>> &tree, int S, int start, 
    vector<vector<int>> &paths, vector<int> &weight)
{
    stack<int> st;
    /*记录累加权重*/
    int sum(0);
    DFS_visit(tree, S, start, paths, st, sum, weight);
}

bool comp(vector<int> v1, vector<int> v2)
{
    if(v1 == v2)
        return false;
    vector<int>::iterator iter_v1 = v1.begin();
    vector<int>::iterator iter_v2 = v2.begin();
    while(1)
        if((*iter_v1) > (*iter_v2))
            return true;
        else if((*iter_v1) < (*iter_v2))
            return false;
        else 
            {++iter_v1; ++iter_v2;}                                                                                                                                                                                                                                                                                                                                                   
}

int main()
{
    int N, M, S;
    while(cin>>N>>M>>S)
    {
        vector<int> colum(N, MAX);
        /*存成有向图*/
        vector<vector<int>> tree(N, colum);
        vector<int> weight(N, 0);
        for(int i=0; i<N; ++i)
            cin>>weight[i];
        for(int i=0; i<M; ++i)
        {
            int id, k; cin>>id>>k;
            for(int j=0; j<k; ++j)
            {
                int n; cin>>n;
                tree[id][n] = 1;
            }
        }
        /*存储结果路径*/
        vector<vector<int>> paths;
        DFS_version(tree, S, 0, paths, weight);
        sort(paths.begin(), paths.end(), comp);
        for(int i=0; i<paths.size(); ++i)
        {
            for(int j=0; j<paths[i].size()-1; ++j)
                cout<<paths[i][j]<<" ";
            cout<<paths[i][paths[i].size()-1]<<endl;
        }
    }
    return 0;
}

题目链接：http://pat.zju.edu.cn/contests/pat-a-practise/1053

思路; 将树存成有向图，用DFS，配一个栈记录路径，最后对多条结果路径排序下，比较简单！