PAT 1053. Path of Equal Weight (30) (dfs + 路径打印）

1053. Path of Equal Weight (30)

时间限制

10 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

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_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

深搜满足条件打印路径的题目，题意是说给了一棵每个节点都有相应权值的树，规定根节点是0，从根节点向叶子节点计算，若一直走到某个叶子节点的所有权值和为给定的数s，就要打印这条路径，print all the paths with weight S in non-increasing order，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.

根据这个打印要求就要对每个非叶子节点的子节点按权值排序，然后满足条件的路径利用pre数组（节点前驱数组）来打印。

#include <cstdio>#include <cstdlib>#include <cstring>#include <map>#include <vector>#include <stack>#include <algorithm>using namespace std;const int MAX = 101;int weigh[MAX];map<int,vector<int> >adjlist;int pre[MAX];stack<int> path;void print(int p){while(p!=-1){path.push(p);p = pre[p];}printf("%d",weigh[path.top()]);path.pop();while(!path.empty()){printf(" %d",weigh[path.top()]);path.pop();}printf("\n");}void dfs(int p,const int s,int sum){sum += weigh[p];if(sum>s)return;if(sum==s && adjlist[p].empty()){//权值和等于s且为叶子，打印路径print(p);return;}//说明到此处的权值和小于s，若p为叶子，则返回，此路无解if(adjlist[p].empty())return;//遍历其各个孩子路径vector<int>::reverse_iterator ite = adjlist[p].rbegin();for(;ite!=adjlist[p].rend();++ite){dfs(*ite,s,sum);}}void init(int n){int i;for(i=0;i<n;++i){pre[i] = -1;}}bool cmp(int a,int b){return weigh[a]<weigh[b];}int main(){//freopen("in.txt","r",stdin);int n,m,s,i,k,id,sid,ln;scanf("%d %d %d",&n,&m,&s);for(i=0;i<n;++i){scanf("%d",&weigh[i]);}init(n);while(m--){scanf("%d %d",&id,&ln);while(ln--){scanf("%d",&sid);adjlist[id].push_back(sid);pre[sid] = id;//记录节点的前驱，打印路径时使用}sort(adjlist[id].begin(),adjlist[id].end(),cmp);}dfs(0,s,0);return 0;}