UVA 12219 Common Subexpression Elimination(STL)

题意就是求最小的表达式树，也就是把相同的表达式子树给替换成最前面相同的编号。

做法：因为需要一一比较，所以把树给编号，方法是用根节点的字符串加左子树编号以及右子树编号。这样经过一次dfs就可以知道哪些是有重复的了。他要输出的编号可以用一个全局变量x，过程中遇到新的字符串就加加，如果这个树与子树是前面出现过的，那么就把x--。结构体里不仅要保存根节点字符串，左右子树编号，还要保存第一次出现的位置，这样才能满足输出的条件。

AC代码：

#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<ctype.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<ctime>
#include<string.h>
#include<string>
#include<sstream>
#include<bitset>
using namespace std;
#define ll __int64
#define ull unsigned long long
#define eps 1e-8
#define NMAX 1000000000
#define MOD 1000000
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1)
#define ALL(x) x.begin(), x.end()
#define INS(x) inserter(x, x.end())
template<class T>
inline void scan_d(T &ret)
{
char c;
int flag = 0;
ret=0;
while(((c=getchar())<'0'||c>'9')&&c!='-');
if(c == '-')
{
flag = 1;
c = getchar();
}
while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();
if(flag) ret = -ret;
}
template<class T> inline T Max(T a, T b){ return a > b ? a : b;}
template<class T> inline T Min(T a, T b){ return a < b ? a : b;}
const int maxn = 50000+10;
char ch[maxn*10];
int lch[maxn],rch[maxn],nct,pos;
char op[maxn][6];
struct node
{
char s[6];
int l,r,wei;
bool operator < (const node &t) const
{
int p = strcmp(s,t.s);
if(p == 0) return l == t.l ? r < t.r : l < t.l;
return p > 0 ? 0 : 1;
}
};
map<node, int>mp;
vector<node>v;
int flag[maxn],ji;
bool pp;
int ID(node x)
{
if(mp.find(x) != mp.end())
{
pp = 1;
return mp[x];
}
v.push_back(x);
return mp[x] = v.size()-1;
}

node make_node(char *s, int l, int r, int wei)
{
node t; strcpy(t.s,s); t.l = l; t.r = r; t.wei = wei; return t;
}

int build(int u)
{
int i = 0;
while(ch[pos] >= 'a' && ch[pos] <= 'z')
{
op[u][i++] = ch[pos++];
}
int qi = ++ji;
op[u][i] = '\0';
//    cout<<op[u]<<" "<<pos<<endl;
if(ch[pos] == '(')
{
pos++;
lch[u] = ++nct;
int ID1 = build(lch[u]);
pos++;
rch[u] = ++nct;
int ID2 = build(rch[u]);
pos++;
pp = 0;
int ret = ID(make_node(op[u],ID1,ID2,qi));
if(pp)
{
flag[u] = v[ret].wei;
ji--;
}
return ret;
}
pp = 0;
int ret = ID(make_node(op[u],-1,-1,qi));
if(pp)
{
flag[u] = v[ret].wei;
ji--;
}
//    if(pos == 21) cout<<op[u]<<" "<<flag[u]<<" "<<qi<<endl;
return ret;
}

void dfs(int u)
{
if(flag[u])
{
printf("%d",flag[u]);
return;
}
printf("%s",op[u]);
if(lch[u] != 0)
{
printf("(");
dfs(lch[u]);
printf(",");
dfs(rch[u]);
printf(")");
}
}

int main()
{
#ifdef GLQ
freopen("input.txt","r",stdin);
//    freopen("o.txt","w",stdout);
#endif
int cas;
scanf("%d\n",&cas);
while(cas--)
{
scanf("%s",ch);
nct = 1; pos = ji = 0;
mp.clear();
v.clear();
memset(flag,0,sizeof(flag));
memset(lch,0,sizeof(lch));
memset(rch,0,sizeof(rch));
build(1);
dfs(1);
printf("\n");
}
return 0;
}