PAT（顶级）2020年秋季考试 7-3 Color the Tree (35分)

7-3 Color the Tree (35分)
 

There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 properties:

  • (1) Every node is either red or black.
  • (2) The root is black.
  • (3) Every leaf (NULL) is black.
  • (4) If a node is red, then both its children are black.
  • (5) For each node, all simple paths from the node to descendant leaves contain the same number of black nodes.

For example, the tree in Figure 1 is a red-black tree, while the ones in Figure 2 and 3 are not.

rbf1.jpgrbf2.jpgrbf3.jpg
Figure 1 Figure 2 Figure 3

For each given binary search tree, you are supposed to tell if it is possible to color the nodes and turn it into a legal red-black tree.

Input Specification:

Each input file contains several test cases. The first line gives a positive integer K (≤10) which is the total number of cases. For each case, the first line gives a positive integer N (≤30), the total number of nodes in the binary search tree. The second line gives the postorder traversal sequence of the tree. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in a line Yes if the given tree can be turned into a legal red-black tree, or No if not.

Sample Input:

3
9
1 4 5 2 8 15 14 11 7
9
1 4 5 8 7 2 15 14 11
8
6 5 8 7 11 17 15 10
 

Sample Output:

Yes
No
Yes

AC代码：
  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 const int maxn=111;
  5 struct node
  6 {
  7     int l,r,tol;
  8 }stu[maxn];
  9 int n,a[maxn],root,tot;
 10 void handle(int l,int r,int pre)
 11 {
 12     if(l==r) return ;
 13     int pos=r;
 14     for(int i=r-1;i>=l;--i) if(a[i]>a[r]) pos=i;
 15     if(pos==r){
 16         int cnt=++tot;
 17         stu[cnt].l=0;
 18         stu[cnt].r=0;
 19         handle(l,pos-1,cnt);
 20         stu[pre].l=cnt;
 21     }
 22     else if(pos==l){
 23         int cnt=++tot;
 24         stu[cnt].l=0;
 25         stu[cnt].r=0;
 26         handle(pos,r-1,cnt);
 27         stu[pre].r=cnt;
 28     }
 29     else{
 30         int cnt=++tot;
 31         stu[cnt].l=0;
 32         stu[cnt].r=0;
 33         handle(l,pos-1,cnt);
 34         stu[pre].l=cnt;
 35         cnt=++tot;
 36         stu[cnt].l=0;
 37         stu[cnt].r=0;
 38         handle(pos,r-1,cnt);
 39         stu[pre].r=cnt;
 40     }
 41 }
 42 vector<int> solve(vector<int> p,vector<int> q)
 43 {
 44     vector<int> res;
 45     res.clear();
 46     for(int i=0;i<(int)p.size();++i){
 47         int u=p[i];
 48         for(int j=0;j<(int)q.size();++j){
 49             if(u==q[j]){
 50                 res.push_back(u);
 51                 break;
 52             }
 53         }
 54     }
 55     return res;
 56 }
 57 int tol;
 58 vector<int> red[maxn],black[maxn];
 59 bool flag;
 60 void dfs(int cur)
 61 {
 62     if(flag==false) return ;
 63     if(stu[cur].l>0 && stu[cur].r>0){
 64         dfs(stu[cur].l);
 65         if(flag==false) return ;
 66         dfs(stu[cur].r);
 67         if(flag==false) return ;
 68         vector<int> x=solve(black[stu[stu[cur].l].tol],black[stu[stu[cur].r].tol]);
 69         int cnt=++tol;
 70         stu[cur].tol=cnt;
 71         red[cnt]=x;
 72         black[cnt].clear();
 73         vector<int> y=solve(black[stu[stu[cur].l].tol],red[stu[stu[cur].r].tol]);
 74         vector<int> z=solve(red[stu[stu[cur].l].tol],black[stu[stu[cur].r].tol]);
 75         vector<int> w=solve(red[stu[stu[cur].l].tol],red[stu[stu[cur].r].tol]);
 76         set<int> se;
 77         se.clear();
 78         for(int i=0;i<(int)x.size();++i) se.insert(x[i]);
 79         for(int i=0;i<(int)y.size();++i) se.insert(y[i]);
 80         for(int i=0;i<(int)z.size();++i) se.insert(z[i]);
 81         for(int i=0;i<(int)w.size();++i) se.insert(w[i]);
 82         set<int>::iterator it;
 83         for(it=se.begin();it!=se.end();++it){
 84             black[cnt].push_back((*it)+1);
 85         }
 86         if((int)red[cnt].size()==0 && (int)black[cnt].size()==0){
 87             flag=false;
 88             return ;
 89         }
 90     }
 91     else if(stu[cur].l>0){
 92         dfs(stu[cur].l);
 93         if(flag==false) return ;
 94         int id=stu[stu[cur].l].tol;
 95         bool have0=false;
 96         for(int i=0;i<(int)red[id].size();++i){
 97             if(red[id][i]==0){
 98                 have0=true;
 99                 break;
100             }
101         }
102         if(have0==false){
103             flag=false;
104             return ;
105         }
106         int cnt=++tol;
107         stu[cur].tol=cnt;
108         red[cnt].clear();
109         black[cnt].clear();
110         black[cnt].push_back(1);
111     }
112     else if(stu[cur].r>0){
113         dfs(stu[cur].r);
114         if(flag==false) return ;
115         int id=stu[stu[cur].r].tol;
116         bool have0=false;
117         for(int i=0;i<(int)red[id].size();++i){
118             if(red[id][i]==0){
119                 have0=true;
120                 break;
121             }
122         }
123         if(have0==false){
124             flag=false;
125             return ;
126         }
127         int cnt=++tol;
128         stu[cur].tol=cnt;
129         red[cnt].clear();
130         black[cnt].clear();
131         black[cnt].push_back(1);
132     }
133     else{
134         int cnt=++tol;
135         stu[cur].tol=cnt;
136         red[cnt].clear();
137         red[cnt].push_back(0);
138         black[cnt].clear();
139         black[cnt].push_back(1);
140     }
141 }
142 int main()
143 {
144     int t;
145     for(scanf("%d",&t);t;--t){
146         scanf("%d",&n);
147         for(int i=1;i<=n;++i){
148             scanf("%d",&a[i]);
149         }
150         tot=0;
151         root=++tot;
152         stu[root].l=0;
153         stu[root].r=0;
154         handle(1,n,1);
155         flag=true;
156         tol=0;
157         dfs(root);
158         if(flag) printf("Yes\n");
159         else printf("No\n");
160     }
161     return 0;
162 }
View Code

 

posted @ 2020-09-22 14:22  lglh  阅读(400)  评论(0)  编辑  收藏  举报