UVA - 656 Optimal Programs

题目大意：给出两个序列，要求找出一个函数 f(x) = y，xi 与 yi 分别对应两个序列中的第 i 个元素，输出函数的操作步骤，要求最短，且按照字典序输出。

解题思路：递归枚举 x0 到 y0 的可能步骤，然后判断其他是否全部满足。

#include <cstdio>
#include <stack>
#include <cmath>
using namespace std;

int T, n, ans, start[15], end[15], rec[20], path[20];
char str[5][5] = {"ADD", "DIV", "DUP", "MUL", "SUB"};
stack <int> S;


void init() {
    ans = 11; 

    for (int i = 0; i < n; i++)
        scanf("%d", &start[i]);

    for (int i = 0; i < n; i++)
        scanf("%d", &end[i]);
}

bool judge(int cur) {
    for (int i = 1; i < n; i++) {
        stack <int> SS;
        SS.push(start[i]);

        for (int j = 0; j < cur; j++) {
            int a = SS.top(), b;
            if (abs(a) > 30000 || (rec[j] == 1 && a == 0))
                return false;

            if (rec[j] != 2) {
                SS.pop();
                b = SS.top(); 
                SS.pop();
            }

            switch (rec[j]) {
                case 0: SS.push(a + b); break;
                case 1: SS.push(b / a); break;
                case 2: SS.push(a); break;
                case 3: SS.push(a * b); break;
                case 4: SS.push(b - a); break;
            }
        }

        if (SS.top() != end[i])
            return false;
    }
    return true;
}

void DFS(int cur, int DUP) {
    if (S.top() == end[0] && ans > cur && judge(cur)) {
        ans = cur;
        for (int i = 0; i < cur; i++)
            path[i] = rec[i];
        return;
    }

    if (cur == ans - 1 || abs(S.top()) > 30000)
        return;

    if (S.size() > 1) {
        int a = S.top(); S.pop();
        int b = S.top(); S.pop();

        rec[cur] = 0;
        S.push(a + b);
        DFS(cur + 1, DUP);
        S.pop();


        if (a != 0) {
            rec[cur] = 1;
            S.push(b / a);
            DFS(cur + 1, DUP);
            S.pop();
        }

        S.push(b);
        S.push(a);
    }

    if (DUP < 5) {
        rec[cur] = 2;
        S.push(S.top());
        DFS(cur + 1, DUP + 1);
        S.pop();
    }

    if (S.size() > 1) {
        int a = S.top(); S.pop();
        int b = S.top(); S.pop();

        rec[cur] = 3;
        S.push(a * b);
        DFS(cur + 1, DUP);
        S.pop();


        rec[cur] = 4;
        S.push(b - a);
        DFS(cur + 1, DUP);
        S.pop();

        S.push(b);
        S.push(a);
    }
}

void put() {
    printf("Program %d\n", ++T);

    if (ans == 0)
        puts("Empty sequence\n");
    else if (ans == 11)
        puts("Impossible\n");
    else {
        for (int i = 0; i < ans - 1; i++)
            printf("%s ", str[path[i]]);
        printf("%s\n\n", str[path[ans-1]]);
    }
}

int main() {
    while (scanf("%d", &n), n) {
        init();

        S.push(start[0]);
        DFS(0, 0);
        S.pop();

        put();
    }
    return 0;
}