1053 Path of Equal Weight (30 分)

Given a non-empty tree with root R, and with weight Wi​​ assigned to each tree node Ti​​. 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 Wi​​ (<) 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 { is said to be greater than sequence { if there exists 1 such that Ai​​=Bi​​ for ,, 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

AC代码:
#include<bits/stdc++.h>
using namespace std;
const int maxn=1010;
#define  inf  0x3fffffff
struct node{
    int w;
    vector<int> child;
};
vector<node> v;
vector<int> path;
int n,m,t;
bool cmp(int a,int b){
    return v[a].w>v[b].w;
}
void dfs(int index,int nodeNum,int sum){
    if(sum>t){
        return;
    }
    if(sum==t){
        if(v[index].child.size()!=0){
            return;
        }
        for(int i=0;i<nodeNum;i++){
            if(i<nodeNum-1){
                printf("%d ",v[path[i]].w);
            }
            else{
                printf("%d
",v[path[i]].w);
            }
            
        }
        return ;
    }
    for(int i=0;i<v[index].child.size();i++){
        int nod=v[index].child[i];
        path[nodeNum]=nod;
        dfs(nod,nodeNum+1,sum+v[nod].w);
    }
}
int main(){
    
    scanf("%d %d %d",&n,&m,&t);
    v.resize(n);
    path.resize(n);
    for(int i=0;i<n;i++){
        scanf("%d",&v[i].w);
    }
    int id,k;
    for(int i=0;i<m;i++){
        scanf("%d %d",&id,&k);
        v[id].child.resize(k);
        for(int j=0;j<k;j++){
            scanf("%d",&v[id].child[j]);
        }
        sort(v[id].child.begin(),v[id].child.end(),cmp);
    }
    dfs(0,1,v[0].w);
    
    

    return 0;
}
原文地址:https://www.cnblogs.com/dreamzj/p/14433124.html