Pat(Advanced Level)Practice--1053(Path of Equal Weight)

Pat1053代码

题目描述：

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 Lis 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

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_ifor 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

AC代码：

对其进行深度优先搜索即可，注意输出的顺序，所以我们这里先对权重较大的节点进行搜索，这样就免去了排序的步骤了。

#include<cstdio>#include<vector>#include<algorithm>#define MAX 105using namespace std;vector<int> adjoin[MAX];vector<int> path;int visited[MAX],weight[MAX];int n,m,s;void DFS(int w,int ID){int len;visited[ID]=1;w+=weight[ID];if(w>s)return;//权重大于S，直接返回else if(w==s&&adjoin[ID].size()==0)//权重等于S，并且为叶子节点{path.push_back(weight[ID]);printf("%d",path[0]);len=path.size();for(int i=1;i<len;i++)printf(" %d",path[i]);printf("\n");path.pop_back();}else if(w<s&&adjoin[ID].size()>0){int maxweight=0,maxid=0;path.push_back(weight[ID]);while(maxid!=-1){     maxweight=maxid=-1;for(int i=0;i<adjoin[ID].size();i++)//从该节点孩子中权{                            //重最大的开始搜索if(!visited[adjoin[ID][i]]&&weight[adjoin[ID][i]]>maxweight){maxweight=weight[adjoin[ID][i]];maxid=adjoin[ID][i];}}if(maxid!=-1){visited[maxid]=1;DFS(w,maxid);}}path.pop_back();}return;}int main(int argc,char *argv[]){int i,j,k;int ID,child;scanf("%d%d%d",&n,&m,&s);for(i=0;i<n;i++)scanf("%d",&weight[i]);for(i=0;i<m;i++){scanf("%d%d",&ID,&k);for(j=0;j<k;j++){scanf("%d",&child);adjoin[ID].push_back(child);}}DFS(0,0);return 0;}