1053. Path of Equal Weight (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 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 W_i (<1000) 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 {A₁, A₂, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., B_m} if there exists 1 <= k < min{n, m} such that A_i = B_i for i=1, ... k, and A_k+1 > B_k+1.

Sample Input:

20 9 2410 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 200 4 01 02 03 0402 1 0504 2 06 0703 3 11 12 1306 1 0907 2 08 1016 1 1513 3 14 16 1717 2 18 19

Sample Output:

10 5 2 710 4 1010 3 3 6 210 3 3 6 2

n为结点数，m为非叶子结点数，S为权重数。

n个结点的权重。

m行父节点和子节点的关系。

如果从根节点到叶子结点的权重和为S，则从大到小输出。

参照算法笔记P304.

1.输入时，对所有子节点的权重从大到小排序，dfs时的输出顺序就是从大到小。

2.好好背dfs模板。

结构体中保存子节点 vector<int>child

#include<stdio.h>#include<vector>#include<algorithm>using namespace std;int S;vector<int>path;struct node{int id;int weight;vector<int>child;}stu[110];bool comp(int a,int b){return stu[a].weight>stu[b].weight;}void dfs(int s){if(stu[s].child.size()==0){path.push_back(s);int dis=0,i;for(i=path.size()-1;i>=0;i--){dis=dis+stu[path[i]].weight;}if(dis==S){printf("%d",stu[path[0]].weight);for(i=1;i<path.size();i++){printf(" %d",stu[path[i]].weight);}printf("\n");}path.pop_back();return;}path.push_back(s);int i;for(i=0;i<stu[s].child.size();i++){dfs(stu[s].child[i]);}path.pop_back();}int main(){int n,m,i,j,id,k,tempid;scanf("%d %d %d",&n,&m,&S);for(i=0;i<n;i++){scanf("%d",&stu[i].weight);stu[i].id=i;}for(i=0;i<m;i++){scanf("%d %d",&id,&k);for(j=0;j<k;j++){scanf("%d",&tempid);stu[id].child.push_back(tempid);}sort(stu[id].child.begin(),stu[id].child.end(),comp);}dfs(0);} 

#include<stdio.h>#include<vector>#include<algorithm>using namespace std;int n,m,s;struct node{int weight;vector<int>child;}tree[110];//bool comp(int a,int b){//return a>b;//}bool comp(int a,int b){return tree[a].weight>tree[b].weight;}vector<int>path;void dfs(int root,int sum){path.push_back(root);if(sum>s){path.pop_back();return;}if(tree[root].child.size()==0&&sum==s){int i;for(i=0;i<path.size();i++){printf("%d",tree[path[i]].weight);if(i!=path.size()-1){printf(" ");}}printf("\n");path.pop_back();return;}int i;for(i=0;i<tree[root].child.size();i++){int next=tree[root].child[i];dfs(next,sum+tree[next].weight);}path.pop_back();//!!!!!!!}int main(){int i,j,k,weight,father,child;scanf("%d %d %d",&n,&m,&s);for(i=0;i<n;i++){scanf("%d",&tree[i].weight);}for(i=0;i<m;i++){scanf("%d %d",&father,&k);for(j=0;j<k;j++){scanf("%d",&child);tree[father].child.push_back(child);}sort(tree[father].child.begin(),tree[father].child.end(),comp);}dfs(0,tree[0].weight);}