Pat(Advanced Level)Practice--1053(Path of Equal Weight)
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
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
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 19Sample Output:
10 5 2 7 10 4 10 10 3 3 6 2 10 3 3 6 2AC程式碼： 對其進行深度優先搜尋即可，注意輸出的順序，所以我們這裡先對權重較大的節點進行搜尋，這樣就免去了排序的步驟了。
#include<cstdio>
#include<vector>
#include<algorithm>
#define MAX 105
using 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;//權重大於S，直接返回
else if(w==s&&adjoin[ID].size()==0)//權重等於S，並且為葉子節點
{
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++)//從該節點孩子中權
{ //重最大的開始搜尋
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;
}