ZOJ - 3965 Binary Tree Restoring

传：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3965

题意：

　　给定同一颗树的两个DFS的序列，输出这颗树。

思路：

　　dfs，每次递归记录两个区间l1, r1, l2, r2 和pa.

#include <bits/stdc++.h>

using namespace std;
            const int maxn = 1e5+9;
            int a[maxn],b[maxn];
            int fa[maxn];
            int pta[maxn], ptb[maxn];
            int sz[maxn];
            int n;
            void dfs(int al, int ar, int bl, int br, int pa) {
                if(al > ar || bl > br) return;
                if(al == ar || bl == br) {
                    fa[a[al]] = pa;
                    fa[b[bl]] = pa;
                    sz[pa] ++;
                    return;
                }

                if(a[al] == b[bl]) {
                    fa[a[al]] = pa;
                    sz[pa] ++;
                    dfs(al+1, ar, bl+1, br, a[al]);
                }
                else {
                    fa[a[al]] = pa;
                    fa[b[bl]] = pa;

                    sz[pa] += 2;

                    int l1 = pta[b[bl]] - 1 - al - 1;
                    dfs(al+1, pta[b[bl]] - 1, ptb[a[al]]+1, ptb[a[al]]+1 + l1 ,a[al]);

                    int l2 = ptb[a[al]] - 1 - bl - 1;
                    dfs(pta[b[bl]] + 1, pta[b[bl]] + 1 + l2 , bl+1, ptb[a[al]] - 1, b[bl]);

                    int mxa = al + l1 + l2 + 3;
                    int mxb = bl + l1 + l2 + 3;
                    int u = pa;
                    while(sz[u] >= 2 )  u = fa[u];
                    dfs(mxa+1, ar, mxb+1, br, u);
                }
            }

int main(){
            int T;  scanf("%d", &T);

            while(T--) {
                scanf("%d", &n);
                for(int i=1; i<=n; i++) scanf("%d", &a[i]), pta[a[i]] = i, sz[i] = 0, fa[i] = 0;
                sz[0] = 0;
                for(int i=1; i<=n; i++) scanf("%d", &b[i]), ptb[b[i]] = i;
                dfs(1, n, 1, n, 0);
                for(int i=1; i<n; i++) printf("%d ", fa[i]);
                printf("%d\n", fa[n]);
            }
            return 0;
}

View Code