hdu 4192(表达式求值)

题意：给一个表达式当中有一些变量，然后告诉你一些数字你可以任意排列，问能不能求出要求的结果。

思路：由于变量数目较小所以直接全排列枚举即可，然后用栈处理表达式。

代码如下：

/**************************************************
 * Author     : xiaohao Z
 * Blog     : http://www.cnblogs.com/shu-xiaohao/
 * Last modified : 2014-06-28 21:50
 * Filename     : hdu_4192.cpp
 * Description     : 
 * ************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<unsigned int,unsigned int> puu;
typedef pair<int, double> pid;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;

const int INF = 0x3f3f3f3f;
const double eps = 1E-6;
const int LEN = 10100;
int n, res, a, num[LEN], len;
char str[LEN];
map<char, int> mp;

struct P{
    int x;
    char c;
    P(){}
    P(int _x){x = _x;}
    P(char _c){c = _c;}
};

int top;
int ch(char c){
    if(!mp.count(c)) mp[c] = top++;
    return mp[c];
}

bool calc(){
    stack<P> s;
    for(int i=0; i<len; i++){
        if(str[i] == '(') continue;
        else if(str[i] == '+' || str[i] == '-' || str[i] == '*'){
            s.push(P(str[i]));
        }else if(str[i] >= 'a' && str[i] <= 'z'){
            s.push(P(num[ch(str[i])]));
        }else if(str[i] == ')'){
            P a = s.top();s.pop();
            P op = s.top();s.pop();
            P b = s.top();s.pop();
            switch(op.c){
                case '+': b.x+=a.x;break;
                case '-': b.x-=a.x;break;
                case '*': b.x*=a.x;break;
            }
            s.push(b);
        }
    }
    return s.top().x == res;
}

bool solve(){
    sort(num, num + n);
    do{
        if(calc()) return true;
    }while(next_permutation(num, num + n));
    return false;
}

int main()
{
//    freopen("in.txt", "r", stdin);

    while(scanf("%d", &n)!=EOF){
        mp.clear();top = 0;
        for(int i=0; i<n; i++)
            scanf("%d", &num[i]);
        scanf("%d", &res);
        if(!n && !res) break;
        scanf("%s", &str);
        len = strlen(str);
        if(solve()) puts("YES");
        else puts("NO");
    }
    return 0;
}