ACM学习历程——POJ3295 Tautology(搜索,二叉树)
Description
WFF 'N PROOF is a logic game played with dice. Each die has six faces representing some subset of the possible symbols K, A, N, C, E, p, q, r, s, t. A Well-formed formula (WFF) is any string of these symbols obeying the following rules:
- p, q, r, s, and t are WFFs
- if w is a WFF, Nw is a WFF
- if w and x are WFFs, Kwx, Awx, Cwx, and Ewx are WFFs.
The meaning of a WFF is defined as follows:
- p, q, r, s, and t are logical variables that may take on the value 0 (false) or 1 (true).
- K, A, N, C, E mean and, or, not, implies, and equals as defined in the truth table below.
| Definitions of K, A, N, C, and E |
| w x | Kwx | Awx | Nw | Cwx | Ewx |
| 1 1 | 1 | 1 | 0 | 1 | 1 |
| 1 0 | 0 | 1 | 0 | 0 | 0 |
| 0 1 | 0 | 1 | 1 | 1 | 0 |
| 0 0 | 0 | 0 | 1 | 1 | 1 |
A tautology is a WFF that has value 1 (true) regardless of the values of its variables. For example, ApNp is a tautology because it is true regardless of the value of p. On the other hand, ApNq is not, because it has the value 0 for p=0, q=1.
You must determine whether or not a WFF is a tautology.
Input
Input consists of several test cases. Each test case is a single line containing a WFF with no more than 100 symbols. A line containing 0 follows the last case.
Output
For each test case, output a line containing tautology or not as appropriate.
Sample Input
ApNp
ApNq
0
Sample Output
tautology
not 考虑到运算符最多是二元的,将运算符和变量存进二叉树中,结构体中用一个val值来记录是否是变量,为了提高效率,用一个visit数组来记录用到了哪几个变量。此外在最后进行运算的时候,需要二叉树进行后序遍历。此外输入采用先序遍历。 代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <string>
#define inf 0x3fffffff
#define eps 1e-10 using namespace std; struct node
{
char op;
int val;
node *left;
node *right;
}; bool a[5];
bool visit[5]; int Do(char op, int x, int y)
{
switch (op)
{
case 'K':
return x&&y;
case 'A':
return x||y;
case 'N':
return !x;
case 'C':
return !x || y;
case 'E':
return x == y;
}
} bool Input(node *p)
{
char ch;
ch = getchar();
if (ch == '0')
return 0;
p->op = ch;
p->val = -1;
switch (ch)
{
case 'p':
p->val = 0;
visit[0] = 1;
return 1;
case 'q':
p->val = 1;
visit[1] = 1;
return 1;
case 'r':
p->val = 2;
visit[2] = 1;
return 1;
case 's':
p->val = 3;
visit[3] = 1;
return 1;
case 't':
p->val = 4;
visit[4] = 1;
return 1;
case 'N':
p->left = (node *)malloc(sizeof(node));
return Input(p->left);
default:
p->left = (node *)malloc(sizeof(node));
p->right = (node *)malloc(sizeof(node));
Input(p->left);
return Input(p->right);
}
} bool caculate(node *p)
{
if (p->val != -1)
return a[p->val];
if (p->op == 'N')
return Do(p->op, caculate(p->left), 1);
else
return Do(p->op, caculate(p->left), caculate(p->right));
} bool dfs(int now, node *head)
{
if (now == 5)
return caculate(head);
if (visit[now] == 0)
return dfs(now+1, head);
int ii, jj;
a[now] = 0;
ii = dfs(now+1, head);
a[now] = 1;
jj = dfs(now+1, head);
return ii && jj;
} bool qt(node *head)
{
if (dfs(0, head))
printf("tautology\n");
else
printf("not\n");
} int main()
{
//freopen("test.txt", "r", stdin);
node *head;
for (;;)
{
memset(visit, 0, sizeof(visit));
head = (node *)malloc(sizeof(node));
if(!Input(head))
break;
getchar();
qt(head);
}
return 0;
}
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