HDOJ 4096 / LA 5708 Universal Question Answering System

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4096

题目大意：

给若干如下的论断：

noun_phrase are noun_phrase.

noun_phrase can verb_phrase.

everything which can verb_phrase can verb_phrase.

everything which can verb_phrase are noun_phrase.

再给出若干如下询问：

are noun_phrase noun_phrase?

can noun_phrase verb_phrase?

can everything which can verb_phrase verb_phrase?

are everything which can verb_phrase noun_phrase?

问此询问是否是正确的。

算法:

实际上就是一道输入输出题

把每个名词或者动词当作一个点，注意一个词可能既是名词又是动词。

然后如果一个句子中出现的第一个词是A，第二次是B，就连一条A->B的边。

判断时，如果一个句子中出现的第一个词是A，第二次是B，那么如果存在A->B的路径，结论就是正确的。

需要注意的trick是。可能出现"are A A?"和"can everything which can A A?"的情况，这种情况无论A是否出现过，结论都是正确的。

注：我的代码目前在uva可以AC但是在HDOJ还是TLE，等到我有时间会继续优化，抱歉。

代码如下：

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<set>
#include<map>
#define INF 0x3f3f3f3f
using namespace std;

const int maxn = 2010;
const int maxm = 1010;
string word[6];
map <string, int> hash[2];
bool vis[maxn << 1];
char s[100000];

int head[maxn << 1], nxt[maxm], to[maxm];
int E = 0;

void dfs(int u, int des)
{
vis[u] = true;
if(vis[des])
{
return;
}
for(int i = head[u]; i != -1; i = nxt[i])
{
int v = to[i];
if(! vis[v])
{
dfs(v, des);
}
}
}

void addedge(int x, int y)
{
to[E] = y;
nxt[E] = head[x];
head[x] = E ++;
}

inline int ffind(string ss, int flg)
{
map <string, int> :: iterator it;
it = hash[flg].find(ss);
int x = hash[flg].size();
if(it == hash[flg].end())
{
hash[flg][ss] = x;
}
else
{
x = it -> second;
}
return x + flg * maxn;
}

int main()
{
int cas;
scanf("%d", &cas);
gets(s);
for(int T = 1; T <= cas; T ++)
{
printf("Case #%d:\n", T);
hash[0].clear();
hash[1].clear();
memset(head, -1, sizeof(head));
E = 0;
while(gets(s))
{
int cot = 0;
if(s[strlen(s) - 1] == '!')
{
puts("");
break;
}
string ss = "";
for(int i = 0; s[i]; i ++)
{
if(s[i] == ' ' || s[i] == '.' || s[i] == '?')
{
word[cot ++] = ss;
ss = "";
}
else
{
ss += s[i];
}
}
int x, y;
if(s[strlen(s) - 1] == '.')
{
if(cot == 3)
{
if(word[1][0] == 'a')
{
x = ffind(word[0], 0);
y = ffind(word[2], 0);
}
else
{
x = ffind(word[0], 0);
y = ffind(word[2], 1);
}
}
else
{
if(word[4][0] == 'c')
{
x = ffind(word[3], 1);
y = ffind(word[5], 1);
}
else
{
x = ffind(word[3], 1);
y = ffind(word[5], 0);
}
}
addedge(x, y);
}
else
{
if(cot == 3)
{
if(word[0][0] == 'a')
{
x = ffind(word[1], 0);
y = ffind(word[2], 0);
}
else
{
x = ffind(word[1], 0);
y = ffind(word[2], 1);
}
}
else
{
if(word[0][0] == 'c')
{
x = ffind(word[4], 1);
y = ffind(word[5], 1);
}
else
{
x = ffind(word[4], 1);
y = ffind(word[5], 0);
}
}
memset(vis, 0, sizeof(vis));
dfs(x, y);
if(vis[y])
{
putchar('Y');
}
else
{
putchar('M');
}
}
}
}
return 0;
}