1053 Path of Equal Weight (30) – 活用sort()及注意
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 < 2^30^, 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 {A~1~, A~2~, …, A~n~} is said to be greater than sequence {B~1~, B~2~, …, 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 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
大致题意为:给出一棵树,每个节点都有权重,若从根节点到叶节点路径的权重和为给定的一个数字的话,就输出这条路径上的权重,路径不唯一。(结合图片理解很清晰,Pintia原图基本都挂)
分析:路径搜索问题,此处显然dfs,直接上代码。
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
vector<int> tab[105];
vector<vector<int> >output;
int power[105] = { 0 };
int n, m, s;
void dfs(vector<int> str, int now, int sum) // 深度优先遍历,搜索符合要求路径
{
sum = sum + power[now]; // 进入结点,先计算当前权值和
str.push_back(power[now]); // 结点入队,此为当前路线结点队列
if (sum > s)return;
if (tab[now].size() == 0)
{
if (sum == s)
{
output.push_back(str); // 利用全局变量记录符合要求路线
}
return;
}
for (int i = 0; i < tab[now].size(); i++)
{
dfs(str, tab[now][i], sum);
}
}
bool cmp(vector<int> &a, vector<int> &b) // 排序规则
{
int len = min(a.size(), b.size());
for (int i = 0; i < len; i++)
{
if (a[i] == b[i])continue;
else return a[i] > b[i];
}
return 0; // 此处必须返回0
}
int main()
{
// **********输入及建树部分**********
scanf("%d %d %d", &n, &m, &s);
for (int i = 0; i < n; i++)scanf("%d", &power[i]);
for (int i = 0; i < m; i++)
{
int parent,sons;
scanf("%d %d", &parent, &sons);
for (int j = 0; j < sons; j++)
{
int s;
scanf("%d", &s);
tab[parent].push_back(s);
}
}
// **********寻找符合要求路径部分**********
vector<int> t;
t.push_back(power[0]);
for (int i = 0; i < tab[0].size(); i++)
{
dfs(t, tab[0][i], power[0]);
}
if (tab[0].size() == 0) // 必须特判根节点无子情况,负责测试点2不通过
{
if (power[0] == s)output.push_back(t);
}
// **********按要求输出部分**********
sort(output.begin(), output.end(), cmp);
for (int i = 0; i < output.size(); i++)
{
printf("%d", output[i][0]);
for (int j = 1; j < output[i].size(); j++)
{
printf(" %d", output[i][j]);
}
printf("\n");
}
return 0;
}
由于结点规模在100以下,我利用vector数组存储每个结点的孩子。深度优先遍历提取所有可能路线,再对路线进行排序。
学会活用sort排序,sort中的cmp函数由自己定义,不仅仅可以实现多属性的排序。对于本题,题目中已经严格给出了对路径大小的定义,可以写出两个vector之间的比较关系,实现cmp函数。
注意:可能会遇见一下错误:
return (_Pred(_Left, _Right) // 引发断点 ? (_Pred(_Right, _Left) ? (_DEBUG_ERROR2("invalid comparator", _File, _Line), true) : true) : false);
对于这种错误,我们一定要注意。cmp()函数实质是判断两个元素在何种状况下交换,交换相当于我们在冒泡排序和选择排序中熟悉的swap()函数,当cmp()返回1时执行swap(),返回0时不执行。
而在sort排序时,一定要注意,函数要求对于调用的两个参数交换位置时不能得到相同的true的结果。详细参见:https://zhidao.baidu.com/question/543126728.html
所以:
bool cmp(vector<int> &a, vector<int> &b) // 排序规则 { int len = min(a.size(), b.size()); for (int i = 0; i < len; i++) { if (a[i] == b[i])continue; else return a[i] > b[i]; } return 0; // 此处必须返回0 }
上述代码若返回1,则会报错,因为在两个vector相同的情况下,必定会交换位置,无论是a在前或者是b在前,都会得到true这个结果。
代码学习
- 养成ac题后查看他人代码好习惯,baidu的第一位的又是一个大神,此题仅用了45行。
- 在考虑输出路径这个问题时,选择了先对儿子结点进行从大到小排序,这样先搜索到的结果也一定是从大到小的。因此不用存储路径直接在叶节点输出即可。
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int target;
struct NODE {
int w;
vector<int> child;
};
vector<NODE> v;
vector<int> path;
void dfs(int index, int nodeNum, int sum) {
if(sum > target) return ;
if(sum == target) {
if(v[index].child.size() != 0) return;
for(int i = 0; i < nodeNum; i++)
printf("%d%c", v[path[i]].w, i != nodeNum - 1 ? ' ' : '\n');
return ;
}
for(int i = 0; i < v[index].child.size(); i++) {
int node = v[index].child[i];
path[nodeNum] = node;
dfs(node, nodeNum + 1, sum + v[node].w);
}
}
int cmp1(int a, int b) {
return v[a].w > v[b].w;
}
int main() {
int n, m, node, k;
scanf("%d %d %d", &n, &m, &target);
v.resize(n), path.resize(n);
for(int i = 0; i < n; i++)
scanf("%d", &v[i].w);
for(int i = 0; i < m; i++) {
scanf("%d %d", &node, &k);
v[node].child.resize(k);
for(int j = 0; j < k; j++)
scanf("%d", &v[node].child[j]);
sort(v[node].child.begin(), v[node].child.end(), cmp1);
}
dfs(0, 1, v[0].w);
return 0;
}