uva10410(树还原BFS)

来源：互联网发布：淘宝视频如何上传编辑：程序博客网时间：2024/06/08 02:30

思路：根据一棵树的先序和中序还原整棵树，记录每个节点的儿子节点，然后打印出来。

此类题目是比较典型的树还原的考点。不同与二叉树的是，这个可能是多叉树，不能像二叉树那样简单的进行儿子节点的判断。

// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#include <limits.h>using namespace std;typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;inline int Readint(){char c = getchar();while(!isdigit(c)) c = getchar();int x = 0;while(isdigit(c)){x = x * 10 + c - '0';c = getchar();}return x;}int in[1010],pre[1010];vector<int> G[1010];struct line{int l,r;};int n;void solve(){queue<line> que;line tmp,x;tmp.l=0,tmp.r=n;que.push(tmp);int k=1;while(!que.empty()){tmp = que.front();que.pop();int u=pre[tmp.l];int top=tmp.l+1;if (tmp.r-tmp.l<=1) continue;G[u].push_back(pre[top]);k++;int i;for (i=top+1;i<tmp.r;++i){if (pre[i]==in[k]){G[u].push_back(pre[i]);k++;x.l=top;x.r=i;top=i;que.push(x);}}if (i==tmp.r){line x;x.l=top;x.r=tmp.r;que.push(x);}}}int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);while(scanf("%d",&n)!=EOF){for (int i=1;i<=n;i++)G[i].clear();for (int i=0;i<n;i++)scanf("%d",&in[i]);for (int i=0;i<n;i++)scanf("%d",&pre[i]);solve();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;}

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