Hut 1997 Seven tombs
题目连接：http://openoj.awaysoft.com/JudgeOnline/problem.php?id=1997
表达式树的应用。
本题可以使用中缀表达式转后缀表达式，然后枚举求解。也可以直接构建表达式树进行枚举求解。
关键是寻找最后计算的运算符，然后递归建立表达式树。
具体做法参考刘汝佳《算法竞赛入门经典》198页。
另外练习：next_permutation（）
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <vector>
#include <algorithm>
using namespace std;
#define maxn 105
int lch[2][maxn];
int rch[2][maxn];
string op[2][maxn];
//节点数
int nc = 0;
int f[7] = {1,2,3,4,5,6,7};
//建立表达式树
int build_tree(char *s,int x,int y,int t)
{
int p = 0;
int c1 = -1;
int c2 = -1;
int flag = 0;
int u = 0;
for(int i=x; i<y; i++)
{
if(s[i] == '(' || s[i] == ')' || s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/')
{
flag = 1;
break;
}
}
//没有运算符了
if(flag == 0)
{
u = ++nc;
lch[t][u] = 0;
rch[t][u] = 0;
for(int i=x; i<y; i++)
{
op[t][u].push_back(s[i]);
}
return u;
}
for(int i=x; i<y; i++)
{
switch(s[i])
{
case '(':
p++;
break;
case ')':
p--;
break;
case '+':
case '-':
if(!p) c1 = i;
break;
case '*':
case '/':
if(!p) c2 = i;
break;
}
}
if(c1<0) c1 = c2;
if(c1<0) return build_tree(s,x+1,y-1,t);
u = ++nc;
lch[t][u] = build_tree(s,x,c1,t);
rch[t][u] = build_tree(s,c1+1,y,t);
op[t][u].push_back(s[c1]);
return u;
}
int num(int u,int t)
{
int temp = 1;
int sum = 0;
if(op[t][u][0] == '+')
{
return num(lch[t][u],t) + num(rch[t][u],t);
}
if(op[t][u][0] == '-')
{
return num(lch[t][u],t) - num(rch[t][u],t);
}
if(op[t][u][0] == '*')
{
return num(lch[t][u],t) * num(rch[t][u],t);
}
for(int i=op[t][u].size()-1; i>=0; i--)
{
sum += f[op[t][u][i] - 'A'] * temp;
temp *= 10;
}
return sum;
}
void init()
{
for(int i=0;i<2;i++)
{
for(int j=0;j<maxn;j++)
{
op[i][j].clear();
}
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
char expression[105];
char left_exp[105];
char right_exp[105];
while(scanf(" %s",expression)!=EOF)
{
init();
int len = strlen(expression);
int len1 = 0,len2 = 0;
int flag = 0;
for(int i=0; i<len; i++)
{
if(expression[i] == '=')
{
flag = 1;
continue;
}
if(flag == 0) left_exp[len1++] = expression[i];
else right_exp[len2++] = expression[i];
}
left_exp[len1] = '\0';
right_exp[len2] = '\0';
nc = 0;
int node1 = build_tree(left_exp,0,len1,0);
nc = 0;
int node2 = build_tree(right_exp,0,len2,1);
int lsum,rsum;
do
{
lsum=num(node1,0);
rsum=num(node2,1);
if(lsum==rsum)
break;
}
while(next_permutation(f,f+7));
for(int i=0; i<len; i++)
{
if(expression[i]>='A' && expression[i]<='G')
printf("%d",f[expression[i]-'A']);
else
printf("%c",expression[i]);
}
printf("\n");
}
}