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