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 = 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
代码如下，扣了2分，暂未找到，本以为是因为第一行输入的不是头节点，还多写了10行代码用于层次遍历，但是错误并不是这个点。。应该是一个边界点，在此先放会。。。唉。。。。
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
bool greater_than(vector<long> v1,vector<long> v2){
for(int i=v1.size()-1,j=v2.size()-1;i>=0&&j>=0;){
if(v1[i]==v2[j]) --j,--i;
else if(v1[i]>=v2[j]) return v1>v2;
else return v1<v2;
}
return v1>v2;
}
int main(){
int n,m;
long s;
scanf("%d%d%ld",&n,&m,&s);
vector<long> weight(n);
vector<long> sum(n);
vector<int> non_leaf(n,0);
vector<int> path(n,-1);
vector<int>map[100];
for(int i=0;i<n;++i){
scanf("%ld",&weight[i]);
sum[i]=weight[i];
}
for(int i=0;i<m;++i){
int father,num,child;
scanf("%d%d",&father,&num);
non_leaf[father]=1;
for(int j=0;j<num;++j){
scanf("%d",&child);
map[father].push_back(child);
sum[child]=weight[child]+sum[father];
// printf("%dss\n",sum[child]);
path[child]=father;
}
}
/*
int root;
vector<int>collection(n,-1);
while(1){
int flag=0;
for(int i=0;i<n;++i){
if(collection[i]!=1&&non_leaf[i]==1){
root=i;
flag=1;
break;
}
}
if(flag==0) break;
collection[root]=1;
for(int i=0;i<map[root].size();++i){
sum[map[root][i]]=weight[map[root][i]]+sum[root];
//printf("%d %d %dss\n",root,map[root][i],sum[map[root][i]]);
}
}*/
vector<vector<long>> v_final;
for(int i=0;i<n;++i){
if(sum[i]==s&&non_leaf[i]==0){
int index=i;
vector<long> v;
while(path[index]!=-1){
v.push_back(weight[index]);
index=path[index];
}
v.push_back(weight[index]);
v_final.push_back(v);
}
}
sort(v_final.begin(),v_final.end(),greater_than);
for(int i=v_final.size()-1;i>=0;--i){
for(int j=v_final[i].size()-1;j>=0;--j){
if(j!=0) printf("%ld ",v_final[i][j]);
else printf("%ld\n",v_final[i][j]);
}
}
return 0;
}
