PAT 1053 Path of Equal Weight



1053. Path of Equal Weight (30)

时间限制
10 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

题意：就是让你从树根开始向下找到所有数字和为ans的所有路径，打出这些路径上每个点对应的值得大小，并且路径是按照路径上值的大小排序的

思路：深搜，从根节点开始深搜它相邻的结点，记录到优先队列里面，这样就不虚要排序了，到时候直接输出路径，进行深搜

AC代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>

using namespace std;

int gra[105][105];//每个节点相邻的边是否存在
int val[105];//每个节点的值
int n,m,ans;//n个结点，m个有叶子，ans为路径上的值

int step[205];//记录路径上的值
int vist[105];//当前路径有没有访问过这个点

struct point {//定义优先队列，值大的在前
int value;
int pos;
bool operator < (const point &a) const {
return value < a.value;
}
} p;

void dfs(int sum,int ct,int nod) {
if(sum == ans) {
int tag = 0;
for(int i = 0; i < n; i++) {//判断这个根下面是不是含有叶子
if(gra[nod][i] && !vist[i]) {
tag =1;
break;
}
}
if(!tag) {//没有的话，输出路径上的值
printf("%d",step[0]);
for(int i = 1; i < ct; i++) {
printf(" %d",step[i]);
}
printf("\n");
return;
}
}
if(sum > ans) {
return;
}
priority_queue <point>q;
for(int i = 0; i < n; i++) {//和当前nod相邻并且没有加入该路径
if(gra[nod][i] && !vist[i]) {
p.value = val[i];
p.pos = i;
q.push(p);
}
}
while(!q.empty()) {
p = q.top();
q.pop();
int i = p.pos;
vist[i] = 1;
step[ct] = val[i];
dfs(sum+p.value,ct+1,i);
vist[i] = 0;
}
}

int main() {
cin >> n >> m >> ans;
memset(vist,0,sizeof(vist));
memset(gra,0,sizeof(gra));
for(int i = 0; i < n; i++) {
cin >> val[i];
}
int s,e,k;
while(m--) {
cin >> s >> k;
while(k--) {
cin >> e;
gra[s][e] = 1;
gra[e][s] = 1;
}
}
step[0] = val[0];
vist[0] = 1;
dfs(val[0],1,0);
return 0;
}