这个题目花了我一晚上才调出来，其实也不难，就是在trie树上搜索给定的单词，但是麻烦的一点是考虑的情况非常多，比如模式串可能重复，或者*？连续出现，最开始都没考虑，程序敲出来直接提交一看WA，然后看discuss里面，说模式可能重复我才知道情况比我想的要复杂，重新该了下程序，还是WA，主要是没考虑到情况*后面忘了判断有无*或者？的情况，最后WA了好几次才A的，程序跑了235M,前30呢，哈哈，多亏了某某大牛的一组测试数据。我会在代码后面附上这组数据，希望后面做这个题得不要犯我这钟低级错误。
测试数据：
View Code
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代码：
#include <iostream>
#include <algorithm>
#include <stdio.h>
using namespace std;
const int N=100002;
const int sonNum=28;
struct nodeIdx{
int index;
nodeIdx *link;
nodeIdx(int i=-1,nodeIdx *l=NULL):index(i),link(l){}
};
struct trieNode{
nodeIdx *pIdx;
int idNums;
trieNode *son[sonNum];
};
nodeIdx sIndex[N];
trieNode node[N*7],root;
char pattrn[7],words[30];
int cnt=0,idxCnt=0,num=0,wordLen;
bool used[N];
void insertP(char * p,int idx)
{
int i;
trieNode *tmp=&root;
while('\0' != *p)
{
if(NULL == tmp->son[*p-'a'])
{
tmp->son[*p-'a']=&node[cnt];
for(i=0;i<sonNum;i++)
node[cnt].son[i]=NULL;
node[cnt].idNums=0;
node[cnt].pIdx=NULL;
cnt++;
}
tmp=tmp->son[*p-'a'];
p++;
}
tmp->idNums++;
if(1 < tmp->idNums)
{
sIndex[idxCnt].index=idx;
sIndex[idxCnt].link=tmp->pIdx;
tmp->pIdx=&sIndex[idxCnt];
idxCnt++;
}
else
{
tmp->pIdx=&sIndex[idxCnt];
sIndex[idxCnt].index=idx;
sIndex[idxCnt].link=NULL;
idxCnt++;
}
}
void query(trieNode* curNode,int curIndex)
{
if(curIndex == wordLen)
{
if(curNode->idNums>0)
{
nodeIdx *tmp=curNode->pIdx;
while(NULL != tmp)
{
if(!used[tmp->index])
{
// ans[num++]=tmp->index;
num++;
used[tmp->index]=true;
}
tmp=tmp->link;
}
}
if((NULL != curNode->son[26]) && (0 < curNode->son[26]->idNums))
{
nodeIdx *tmp=curNode->son[26]->pIdx;
while(NULL != tmp)
{
if(!used[tmp->index])
{
// ans[num++]=tmp->index;
num++;
used[tmp->index]=true;
}
tmp=tmp->link;
}
}
return;
}
if(NULL != curNode->son[words[curIndex]-'a'])
{
query(curNode->son[words[curIndex]-'a'],curIndex+1);
}
if(NULL != curNode->son[26])
{
if(0 < curNode->son[26]->idNums)
{
nodeIdx *tmp=curNode->son[26]->pIdx;
while(NULL != tmp)
{
if(!used[tmp->index])
{
// ans[num++]=tmp->index;
num++;
used[tmp->index]=true;
}
tmp=tmp->link;
}
}
trieNode *tmp=curNode->son[26];
int pos=wordLen-1;
while(pos>=curIndex)
{
if(NULL != tmp->son[words[pos]-'a'])
{
query(tmp->son[words[pos]-'a'],pos+1);
}
//一开始忘了判断*后面是否是？或者*了，贡献了好几个WA啊，考虑不周全
if(NULL != tmp->son[27])
{
query(tmp->son[27],pos+1);
}
if(NULL != tmp->son[26])
{
query(tmp->son[26],pos+1);
}
pos--;
}
}
if(NULL != curNode->son[27])
{
query(curNode->son[27],curIndex+1);
}
}
int main()
{
int n,m,i,j,k;
scanf("%d%d",&n,&m);
root=node[cnt];
for(i=0;i<sonNum;i++)
node[cnt].son[i]=NULL;
node[cnt].idNums=0;
node[cnt].pIdx=NULL;
cnt=1;
for(k=0;k<n;k++)
{
memset(pattrn,'\0',sizeof(pattrn));
scanf("%s",pattrn);
j=strlen(pattrn);
if('*' == pattrn[j-1])
{
//将末尾全是*的缩为一个，多了也没用
while('*' == pattrn[j-2])
{
pattrn[j-1]='\0';
--j;
}
}
for(i=0;i<j;i++)
{
if('*' == pattrn[i])
pattrn[i]='z'+1;
else if('?' == pattrn[i])
pattrn[i]='z'+2;
}
insertP(pattrn,k);
}
for(i=0;i<m;i++)
{
memset(words,'\0',sizeof(words));
// scanf("%s",words);
cin>>words;
num=0;
memset(used,0,sizeof(used));
wordLen=strlen(words);
query(&root,0);
if(num)
{
// qsort(ans,num,sizeof(int),cmp);
for(k=0;k<n;k++)
{
if(used[k])
printf("%d ",k);
}
printf("\n");
// printf("%d",ans[0]);
// for(k=1;k<num;k++)
// printf(" %d",ans[k]);
// printf("\n");
}
else
printf("Not match\n");
}
return 0;
}