UVA 548 Tree

题目大意

　　多组数据，每组数据输入一个二叉树的中序和后序遍历，请你输出一个叶子节点编号，该叶子节点到根的路径上所经过的所有节点编号数值总和最小，且这个叶子是编号最小的那个。编号不重。

分析

　　树结构基础题，不过众所周知，UVA输入输出比较变态。

代码如下

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
  9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 12  
 13 #define pr(x) cout << #x << " = " << x << "  "
 14 #define prln(x) cout << #x << " = " << x << endl
 15  
 16 #define LOWBIT(x) ((x)&(-x))
 17  
 18 #define ALL(x) x.begin(),x.end()
 19 #define INS(x) inserter(x,x.begin())
 20  
 21 #define ms0(a) memset(a,0,sizeof(a))
 22 #define msI(a) memset(a,inf,sizeof(a))
 23 #define msM(a) memset(a,-1,sizeof(a))
 24 
 25 #define MP make_pair
 26 #define PB push_back
 27 #define ft first
 28 #define sd second
 29  
 30 template<typename T1, typename T2>
 31 istream &operator>>(istream &in, pair<T1, T2> &p) {
 32     in >> p.first >> p.second;
 33     return in;
 34 }
 35  
 36 template<typename T>
 37 istream &operator>>(istream &in, vector<T> &v) {
 38     for (auto &x: v)
 39         in >> x;
 40     return in;
 41 }
 42  
 43 template<typename T1, typename T2>
 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 45     out << "[" << p.first << ", " << p.second << "]" << "
";
 46     return out;
 47 }
 48 
 49 inline int gc(){
 50     static const int BUF = 1e7;
 51     static char buf[BUF], *bg = buf + BUF, *ed = bg;
 52     
 53     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
 54     return *bg++;
 55 } 
 56 
 57 inline int ri(){
 58     int x = 0, f = 1, c = gc();
 59     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
 60     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
 61     return x*f;
 62 }
 63  
 64 typedef long long LL;
 65 typedef unsigned long long uLL;
 66 typedef pair< double, double > PDD;
 67 typedef pair< int, int > PII;
 68 typedef pair< int, PII > PIPII;
 69 typedef pair< string, int > PSI;
 70 typedef pair< int, PSI > PIPSI;
 71 typedef set< int > SI;
 72 typedef vector< int > VI;
 73 typedef vector< VI > VVI;
 74 typedef vector< PII > VPII;
 75 typedef map< int, int > MII;
 76 typedef map< int, PII > MIPII;
 77 typedef map< string, int > MSI;
 78 typedef multimap< int, int > MMII;
 79 typedef unordered_map< int, int > uMII;
 80 typedef pair< LL, LL > PLL;
 81 typedef vector< LL > VL;
 82 typedef vector< VL > VVL;
 83 typedef priority_queue< int > PQIMax;
 84 typedef priority_queue< int, VI, greater< int > > PQIMin;
 85 const double EPS = 1e-10;
 86 const LL inf = 0x7fffffff;
 87 const LL infLL = 0x7fffffffffffffffLL;
 88 const LL mod = 1e9 + 7;
 89 const int maxN = 1e4 + 7;
 90 const LL ONE = 1;
 91 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 92 const LL oddBits = 0x5555555555555555;
 93 
 94 struct Node{
 95     int l, r;
 96     int w;
 97 };
 98 
 99 int inorder[maxN], postorder[maxN];
100 int sz;
101 string line;
102 
103 Node tree[maxN];
104 int tlen;
105 
106 int ansSum, ans;
107 
108 void build(int in, int post, int len, int rt) {
109     if(len <= 0) return;
110     tree[rt].w = postorder[post + len - 1];
111     tree[rt].l = tree[rt].r = 0;
112     
113     int tmp = find(inorder, inorder + sz, tree[rt].w) - &inorder[in];
114     if(tmp > 0) {
115         tree[rt].l = ++tlen;
116         build(in, post, tmp, tlen);
117     } 
118     if(len - tmp - 1 > 0) {
119         tree[rt].r = ++tlen;
120         build(in + tmp + 1, post + tmp, len - tmp - 1, tlen);
121     }
122 }
123 
124 bool isLeaf(int rt) {
125     return tree[rt].l == 0 && tree[rt].r == 0;
126 }
127 
128 void solve(int rt, int ret) {
129     ret += tree[rt].w;
130     if(isLeaf(rt)) {
131         if(ansSum == ret) {
132             if(ans > ret) ans = tree[rt].w;
133         }
134         else if(ansSum > ret) {
135             ansSum = ret;
136             ans = tree[rt].w;
137         }
138     }
139     else {
140         if(tree[rt].l) solve(tree[rt].l, ret);
141         if(tree[rt].r) solve(tree[rt].r, ret);
142     }
143 }
144 
145 
146 int main(){
147     //freopen("MyOutput.txt","w",stdout);
148     INIT();
149     while(getline(cin, line)) {
150         ms0(tree);
151         tlen = sz = 0;
152         ansSum = ans = inf;
153         stringstream ss(line);
154         int x;
155         while(ss >> x) inorder[sz++] = x;
156         
157         sz = 0;
158         getline(cin, line);
159         ss.clear();
160         ss.str(line);
161         while(ss >> x) postorder[sz++] = x;
162         
163         build(0, 0, sz, ++tlen);
164         solve(1, 0);
165         cout << ans << endl;
166     }
167     return 0;
168 }

View Code

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原文地址：https://www.cnblogs.com/zaq19970105/p/10978922.html