1053. Path of Equal Weight (30)

时间限制

10 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

遍历，排序。由于时间太短，有时候提交会抽风，超时，多交几次就好了

#include <cstdio>#include <vector>#include <algorithm>#include <iterator>using namespace std;struct point{int weight;int orig_w;int son;int parent;vector <int> sons;};vector <point> p; vector <vector<int>> v;vector <int> group;int weight = 0, s;void tranvers(int root){group.push_back(p[root].orig_w);weight += p[root].orig_w;if(p[root].son == 0 && weight == s){v.push_back(group);}for(int i=0;i<p[root].son;i++){tranvers(p[root].sons[i]);}weight -= p[root].orig_w;group.pop_back();return ;}bool cmp(vector<int> a, vector<int>b){vector<int>::iterator ita = a.begin(),itb = b.begin();while(ita != a.end() && itb != b.end()){if(*ita > *itb)return true;else if(*ita < *itb)return false;ita++; itb++;}return false;}int main(){int n,m,temp,temp2,temp3;scanf("%d%d%d",&n,&m,&s);p.resize(n);for(int i=0;i<n;i++){scanf("%d",&temp);p[i].weight = p[i].orig_w = temp;p[i].son = 0;p[i].parent = -1;}for(int i=0;i<m;i++){scanf("%d%d",&temp,&temp2);p[temp].son = temp2;for(int j=0;j<temp2;j++){scanf("%d",&temp3);p[temp].sons.push_back(temp3);p[temp3].parent = temp;}}tranvers(0);sort(v.begin(),v.end(),cmp);int len = v.size();for(int i=0;i<len;i++){vector<int>::iterator it = v[i].begin();for(it;it!=v[i].end();it++){if(it != v[i].begin())printf(" ");printf("%d",*it);}printf("\n");}return 0;}