题目链接:uva 10410 - Tree Reconstruction
题目大意:给定一个树的BFS和DFS,求这棵树。
解题思路:用栈维护即可。对应BFS序列映射出了每个节点和根节点的距离,遍历dfs序列,对当前节点和栈顶节点比较,如果该节点距离根节点更远,则说明该节点为栈顶节点个孩子节点,则记录后将节点放入栈中。否则弹掉栈顶元素继续比较。需要注意一点,即当元素与栈顶元素的距离值大1的时候要视为相等,因为它们属于兄弟节点
#include <cstdio>
#include <cstring>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 1005;
int N, pos[maxn];
vector<int> g[maxn];
int main () {
while (scanf("%d", &N) == 1) {
int x;
for (int i = 1; i <= N; i++) {
scanf("%d", &x);
g[i].clear();
pos[x] = i;
}
int root;
stack<int> sta;
scanf("%d", &root);
sta.push(root);
for (int i = 1; i < N; i++) {
scanf("%d", &x);
while (true) {
int u = sta.top();
if (u == root || pos[u] + 1 < pos[x]) {
g[u].push_back(x);
sta.push(x);
break;
} else
sta.pop();
}
}
for (int i = 1; i <= N; i++) {
printf("%d:", i);
for (int j = 0; j < g[i].size(); j++)
printf(" %d", g[i][j]);
printf("\n");
}
}
return 0;
}
时间: 2024-10-14 04:27:11