PAT (Advanced) 1053. Path of Equal Weight (30)

原题：1053. Path of Equal Weight (30)

解题思路：
用dfs遍历每一条路径，符合条件输出即可。
为满足输出要求，要对每个非叶结点的子结点按权重从大刀小排序。

代码如下：#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 100 + 5;
int weight[maxn], vis[maxn];
vector<int> G[maxn];

bool cmp(int a, int b)
{
return weight[a] > weight[b];
}
vector<int> ans;
void dfs(int u, int w, int& s)
{
vis[u] = 1;
if(!G[u].size() && w + weight[u] == s) //输出条件
{
ans.push_back(weight[u]);
for(int i = 0; i < ans.size(); i++)
if(i) printf(" %d", ans[i]);
else printf("%d", ans[i]);
printf("\n");
ans.pop_back();
return ;
}
sort(G[u].begin(), G[u].end(), cmp);
ans.push_back(weight[u]);
for(int i = 0; i < G[u].size(); i++)
if(!vis[G[u][i]]) dfs(G[u][i], w+weight[u], s);
ans.pop_back();
}

int main()
{
int n, m, s;
while(scanf("%d%d%d", &n, &m, &s) == 3)
{
for(int i = 0; i < n; i++) scanf("%d", &weight[i]);
for(int i = 0; i < n; i++) G[i].clear();
for(int i = 0; i < m; i++)
{
int num, k, a;
scanf("%d%d", &num, &k);
for(int j = 0; j < k; j++)
{
scanf("%d", &a);
G[num].push_back(a);
}
}
memset(vis, 0, sizeof(vis));
dfs(0, 0, s);
}
return 0;
}