题意
给出五个操作,和n个栈,问能否用同一种操作使得栈中最后的数字,等于目标所要的数字,且栈中任意时刻,每个数字不能超过30000。
输出最小的操作数,如果操作数相同,输出字典序最小的。
解析:
可以先bfs第一个数字和目标第一个数字,然后利用这些操作,操作后面的数字,判断结果是否和目标数字一一对应。
AC代码
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 11;
char str[][5] = {"ADD", "DIV", "DUP", "MUL", "SUB"};
enum oper_set {ADD, DIV, DUP, MUL, SUB};
struct Node {
int step, path[N];
stack<int> st;
Node() {step = 0;}
Node trans(int oper) { //5 operator
Node rhs = *this;
int a, b;
if(oper == DUP) {
a = rhs.st.top();
rhs.st.push(a);
}else {
a = rhs.st.top(); rhs.st.pop();
b = rhs.st.top(); rhs.st.pop();
if(oper == ADD) {
rhs.st.push(a+b);
}else if(oper == DIV) {
rhs.st.push(b/a);
}else if(oper == MUL) {
rhs.st.push(a*b);
}else if(oper == SUB) {
rhs.st.push(b-a);
}
}
return rhs;
}
};
int x[N], y[N], n, ansPath[N];
bool judge(Node rhs) {
for(int i = 1; i < n; i++) {
Node tmp;
tmp.st.push(x[i]);
for(int j = 0; j < rhs.step; j++) {
if(tmp.st.top() == 0 && rhs.path[j] == DIV) return false;
if(abs(tmp.st.top()) > 30000) return false;
tmp = tmp.trans(rhs.path[j]);
}
if(tmp.st.top() != y[i]) return false;
}
return true;
}
int bfs(int star, int end) {
queue<Node> que;
Node begin;
begin.st.push(star); begin.step = 0;
que.push(begin);
while(!que.empty()) {
Node front = que.front(); que.pop();
if(front.st.size() == 1 && front.st.top() == end) {
if(judge(front)) {
memcpy(ansPath, front.path, sizeof(ansPath));
return front.step;
}
}
for(int i = 0; i < 5; i++) {
if(front.st.size() == 1 && i != DUP) continue;
if(front.st.top() == 0 && i == DIV) continue;
Node tmp = front.trans(i);
tmp.path[tmp.step++] = i;
if(abs(tmp.st.top()) > 30000 || tmp.step > 10) continue;
que.push(tmp);
}
}
return INF;
}
int main() {
//freopen("in.txt", "r", stdin);
int cas = 1;
while(scanf("%d", &n) != EOF && n) {
memset(x, 0, sizeof(x));
memset(y, 0, sizeof(y));
for(int i = 0; i < n; i++)
scanf("%d", &x[i]);
for(int i = 0; i < n; i++)
scanf("%d", &y[i]);
printf("Program %d\n", cas++);
if(!memcmp(x, y, sizeof(x))) {
puts("Empty sequence\n");
continue;
}
int rec = bfs(x[0], y[0]);
if(rec == INF) {
puts("Impossible\n");
}else {
printf("%s", str[ansPath[0]]);
for(int i = 1; i < rec; i++) {
printf(" %s", str[ansPath[i]]);
}puts("\n");
}
}
return 0;
}