PAT 1053. Path of Equal Weight (30)

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 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 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 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 = Bifor 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

纸老虎。挺简单的。只要把每个node里的孩子，按降序排一下就好。

#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;

int value[101];

struct
{
int cnt=0;
int child[100];

}node[101];

void paixu(int x)
{
int cnt=node[x].cnt;
if(cnt==0) return;
for(int i=1;i<cnt;i++)
for(int j=0;j<cnt-i;j++)
if(value[node[x].child[j]]<value[node[x].child[j+1]])
swap(node[x].child[j],node[x].child[j+1]);
}

int path[101];
int pp=0;
long long sum=0;
int n,m,s;

void dfs(int x)
{
sum+=value[x];
path[pp++]=value[x];
if(sum==s)
{   if(node[x].cnt==0)
{  int flag=0;
for(int i=0;i<pp;i++)
{
if(flag==1) cout<<' ';
flag=1;
cout<<path[i];
}
cout<<endl;  }
pp--;
sum-=value[x];
return;

}
if(sum>s)
{
pp--;
sum-=value[x];
return;
}
for(int i=0;i<node[x].cnt;i++)
dfs(node[x].child[i]);
pp--;
sum-=value[x];
}

int main()
{

cin>>n>>m>>s;
for(int i=0;i<n;i++)
cin>>value[i];
for(int i=0;i<m;i++)
{
int root,cnt;
cin>>root>>cnt;
node[root].cnt=cnt;
for(int j=0;j<cnt;j++)
cin>>node[root].child[j];
}
for(int i=0;i<n;i++)
paixu(i);
dfs(0);

return 0;
}