UVA - 712 S-Trees（S树）

题意：0往左走，1往右走，已知所有叶子的值，每个查询都是根结点到叶子结点的路径，路径的每一个点分别对应着x1，x2，x3……但是实际上的S树的路径可能并非是x1，x2，x3……

分析：先存路径变量的顺序，来控制最后访问的顺序。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {, , -, , -, -, , };
const int dc[] = {-, , , , -, , -, };
const int MOD = 1e9 + ;
const double pi = acos(-1.0);
const double eps = 1e-;
const int MAXN =  + ;
const int MAXT =  + ;
using namespace std;
char s[MAXN];
char t[MAXN];
vector<int> order;
vector<char> v[];
int POW[];
void init(){//2的i次方
    int tmp = ;
    for(int i = ; i <= ; ++i){
        tmp *= ;
        POW[i] = tmp;
    }
}
int main(){
    int n;
    int kase = ;
    init();
    while(scanf("%d", &n) == ){
        if(n == ) return ;
        for(int i = ; i < ; ++i) v[i].clear();
        order.clear();
        for(int i = ; i < n; ++i){
            char tmp[];
            scanf("%s", tmp);
            order.push_back(tmp[] - '');//S树的变量顺序
        }
        scanf("%s", s);
        int m;
        scanf("%d", &m);
        for(int k = ; k < m; ++k){
            scanf("%s", t);
            for(int i = ; i < n; ++i){
                v[i + ].push_back(t[i] - '');//存x[i+1]的m次查询的值
            }
        }
        printf("S-Tree #%d:\n", ++kase);
        for(int i = ; i < m; ++i){
            int cnt = ;//根结点标号为1
            for(int j = ; j < n; ++j){
                 if(!v[order[j]][i]){//若为0，则2i；否则2i+1
                    cnt *= ;
                 }
                 else{
                    cnt = cnt *  + ;
                 }
            }
            printf("%c", s[cnt - POW[n]]);
        }
        printf("\n\n");
    }
    return ;
}