UVA 548 Tree
作者：互联网
题目链接：https://vjudge.net/problem/UVA-548
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 << "]" << "\n";
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
标签：rt,typedef,const,int,tree,LL,Tree,UVA,548 来源： https://www.cnblogs.com/zaq19970105/p/10978922.html