UVA 12219 Common Subexpression Elimination （dfs瞎搞）

大体题意：

给你一个字符串按照二叉树的形式，用消除公共表达式的方法可以减少表达式树上的的结点，输出最少的结点的图，详细见原题。

思路：

写的比较乱，感觉时间还行吧。借鉴一下吧。

先写dfs 建树，在写个dfs2 从叶子结点向上更新父结点 重新标号，使得相同类的结点归为一类。

在写个dfs3，重新从根节点标号，变得有序。

最后print 函数  要么输出数字 要么输出字符串 讨论一下即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <unordered_map>
#include <string>
#include <map>
#include <iostream>
#define mr make_pair
#define fi first
#define se second
using namespace std;

const int maxn = 50000 + 7;
char s[maxn*20];
struct Node{
string t;
int fii,see;
Node(string t= "", int fii = -1,int see = -1):t(t),fii(fii),see(see){}
bool operator < (const Node& rhs) const {
return fii < rhs.fii || (fii==rhs.fii && see < rhs.see) || (fii==rhs.fii && see == rhs.see && t < rhs.t);
}
bool operator == (const Node& rhs) const {
return t == rhs.t && fii == rhs.fii && see == rhs.see;
}
};

unordered_map<int,bool> mp;
map< Node ,int >Hash;

string node[maxn];
int lch[maxn], rch[maxn],hash_id[maxn],tab[maxn],ord[maxn];
int id, pos,cur,len;
string t;

void dfs(int c){
if (cur >= len) return;
t = s[cur];
while(cur + 1 < len && s[cur+1] != '(' && s[cur+1] != ')' && s[cur+1] != ','){
++cur;
t+= s[cur];
}
node[c] = t;
cur++;
char ch = s[cur];
if (ch == ',') return;
if (ch == ')' || cur >= len ) {
++cur;
return ;
}
if (cur < len && ch == '('){
id++;
lch[c] = id;
++cur;
dfs(id);
}
ch = s[cur];
if (cur < len && ch == ','){
id++;
rch[c] = id;
++cur;
dfs(id);
}
ch = s[cur];

if (cur < len && ch == ')'){
++cur;
return;
}
}

void print(int c){
if (c == -1) return ;

if (mp.count(tab[c])){
printf("%d",tab[c]);
return;
}

printf("%s",node[c].c_str());
mp[tab[c]] = 1;

bool ok = !(lch[c] == -1 && rch[c] == -1);
bool ok2 = !(lch[c] == -1 || rch[c] == -1);

if (ok)printf("(");
print(lch[c]);
if (ok2)printf(",");
print(rch[c]);
if (ok)printf(")");
}
int fii,see;
string tmp;
void dfs2(int cur){
if (cur == -1) return;
dfs2(lch[cur]);
dfs2(rch[cur]);

if (lch[cur] == -1 && rch[cur] == -1){
Node u = Node(node[cur],-1,-1);
if (!Hash.count(u) ){
hash_id[cur] = id++;
Hash[u] = id-1;
}
else {
hash_id[cur] = Hash[u];
}

return;
}
if (lch[cur] == -1) fii = -1;
else fii = hash_id[lch[cur]];

if (rch[cur] == -1) see = -1;
else see = hash_id[rch[cur]];

tmp = node[cur];
Node u = Node(tmp,fii,see);

if (!Hash.count(u) ){
hash_id[cur] = id++;
Hash[u] = id-1;
}
else {
hash_id[cur] = Hash[u];
}

}

void dfs3(int c){
if (c == -1) return;
if (ord[hash_id[c] ] == -1){
tab[c] = id++;
ord[hash_id[c]] = id-1;
}
else {
tab[c] = ord[hash_id[c]];
}
// printf("^^ %d\n",tab[c]);
dfs3(lch[c]);
dfs3(rch[c]);
}

int main(){

int T;
scanf("%d%*c",&T);

while(T--){
memset(lch,-1,sizeof lch);
memset(rch,-1,sizeof rch);
memset(ord,-1,sizeof ord);
len = 0;
scanf("%s",s);
len = strlen(s);
id = pos = cur = 0;
mp.clear();
Hash.clear();
dfs(0);
id = 1;
dfs2(0);

id = 1;
dfs3(0);

print(0);
puts("");
// printf("&&& %d\n",mp[3]);

}

return 0;
}
/**
3
this(is(a,tiny),tree)
a(b(f(a,a),b(f(a,a),f)),f(b(f(a,a),b(f(a,a),f)),f))
z(zz(zzzz(zz,z),zzzz(zz,z)),zzzz(zz(zzzz(zz,z),zzzz(zz,z)),z))

1
a(b(f(a,a),b(f(a,a),f)),f(b(f(a,a),b(f(a,a),f)),f))

**/