poj 1816 trie树+dfs(强烈推荐，含有通配符的匹配)

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//                 printf(" %d",ans[k]);
//            printf("\n");
        }
else
            printf("Not match\n");
    }
return 0;
}

转载于:https://www.cnblogs.com/buptLizer/archive/2011/09/18/2180723.html