1053 Path of Equal Weight (30 分)

1053 Path of Equal Weight (30 分)

 

Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weight 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 the following figure: 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 the figure.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0, the number of nodes in a tree, M (<), the number of non-leaf nodes, and 0, the given weight number. The next line contains N positive numbers where W​i​​ (<) corresponds to the tree node T​i​​. 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 { is said to be greater than sequence { if there exists 1 such that A​i​​=B​i​​ for ,, and A​k+1​​>B​k+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


这题挺不错的，但是算是一个dfs遍历的模板题，没加啥东西。

#include <bits/stdc++.h>
#define ll long long int
using namespace std;
int n,m,k;
struct Node
{
    int to;
    ll val;
    friend bool operator < (const Node &a, const Node &b){
        return a.val > b.val;
    }
};
vector<Node> v[105];
int vis[105];
ll an[105];
int dp[105];


void output(int x){
    stack<int> st;
    while(x != 0){
        st.push(an[x]);
        x = dp[x];
    }
    st.push(an[0]);
    while(!st.empty()){
        cout << st.top();
        if(st.size() == 1){
            cout << endl;
        }else{
            cout << " ";
        }
        st.pop();
    }
}

void dfs(int x, ll sum){
    vis[x] = 1;
    if(sum == k){
        if(v[x].size() == 0)
            output(x);
        return ;
    }else if(sum > k){
        return ;
    }
    for(int i = 0; i < v[x].size(); i++){
        if(vis[v[x][i].to])
            continue;
        dp[v[x][i].to] = x;
        dfs(v[x][i].to, sum+v[x][i].val);
    }
}


int main(){
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 0; i < n; i++){
        cin >> an[i];
    }
    int x,y,z;
    for(int i = 0; i < m; i++){
        scanf("%d%d", &x,&y);
        for(int j = 0; j < y; j++){
            scanf("%d", &z);
            v[x].push_back({z,an[z]});
        }
    }
    for(int i = 0; i < n; i++){
        sort(v[i].begin(), v[i].end());
    }
    dfs(0, an[0]);
    return 0;
}