dfs_1053 Path of Equal Weight (30 分)
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<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 Wi (<1000) 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 {A1,A2,⋯,An} is said to be greater than sequence {B1,B2,⋯,Bm} if there exists 1≤k<min{n,m} such that Ai=Bi for i=1,⋯,k, 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
題目中給出每個節點的權重,以及和每個節點相連的兒子節點,要求求出到達葉子節點上路徑權值為s的所有路徑,將路徑上從根節點到兒子節點經歷的權值按照由大到小的順序輸出,權值的比較為:如果位置上數值相同比較下一位,不相同那麼較大的那個比小的大
根節點固定都為00
我們由根節點向下遍歷,遍歷順序為由大根據兒子節點的權值大小向下比較,首先選擇兒子節點權值大的,後面選擇權值小的
知道葉子節點,如果到達葉子節點權值剛好為s,則輸出這條路徑上的權值,如果在路徑中出現超過s的情況,刪除這種情況,或者沒有到達葉子節點就已經等於s,也刪除
路徑的儲存使用per陣列
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <cmath>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 105;
int n,m,s,w[maxn],per[maxn],kid[maxn];
vector<int> ch[maxn];
bool cmp(int a,int b)
{
return w[a] < w[b];
}
void dfs(int p,int ww)
{
if(ww > s) return ;
if(ww == s && ch[p].size() == 0)
{
int path[maxn],index = 0;
while(p != per[p])
{
path[index++] = p;
p = per[p];
}
path[index++] = p;
sort(path,path+index);
for(int i = 0;i < index;i ++)
printf("%d%c",w[path[i]],i==index-1?'\n':' ');
return ;
}
for(int i = ch[p].size()-1;i >= 0;i --)
{
int x = ch[p][i];
per[x] = p;
dfs(x,ww+w[x]);
per[x] = x;
}
}
int main()
{
scanf("%d%d%d",&n,&m,&s);
for(int i =0 ;i < n;i ++)
scanf("%d",w+i);
for(int i =0 ;i < n;i ++) per[i] = i;
int x,y,z;
for(int i =0 ;i < m;i ++)
{
scanf("%d%d",&x,&y);
for(int j = 0;j < y;j ++)
scanf("%d",kid+j);
sort(kid,kid+y,cmp);
for(int j =0;j < y;j ++)
ch[x].push_back(kid[j]);
}
dfs(0,w[0]);
return 0;
}