1032 Sharing （25 分）

【题目描述】

To store English words, one method is to use linked lists and store a word letter by letter. To save some space, we may let the words share the same sublist if they share the same suffix. For example, loading and being are stored as showed in Figure 1.

Figure 1
You are supposed to find the starting position of the common suffix (e.g. the position of i in Figure 1).

【Input Specification】

Each input file contains one test case. For each case, the first line contains two addresses of nodes and a positive N (≤10​5​​), where the two addresses are the addresses of the first nodes of the two words, and N is the total number of nodes. The address of a node is a 5-digit positive integer, and NULL is represented by −1.
Then N lines follow, each describes a node in the format:
Address Data Next
whereAddress is the position of the node, Data is the letter contained by this node which is an English letter chosen from { a-z, A-Z }, and Next is the position of the next node.

【Output Specification】

For each case, simply output the 5-digit starting position of the common suffix. If the two words have no common suffix, output -1 instead.
【Sample Input 1】
11111 22222 9
67890 i 00002
00010 a 12345
00003 g -1
12345 D 67890
00002 n 00003
22222 B 23456
11111 L 00001
23456 e 67890
00001 o 00010

【Sample Output 1】

67890

【Sample Input 2】

00001 00002 4
00001 a 10001
10001 s -1
00002 a 10002
10002 t -1

【Sample Output 2】

-1

【思路】

根据题目描述，是让我们找首个共用节点地址。
一开始我把题意理解成找相同后缀的首个字符的地址（感觉英文翻译出来就是这个意思啊！不明白哪里理解出问题了），所以有了解决方案2，但是后面两个检测点过不去，经过修改后的解决方案2可以AC了，解决方案1比较优秀，可以参考1，2就不用看了。

【解决方案1】

#include<cstdio>
#include<cstring>
const int maxn = 100010;
struct NODE{
    char data;
    int next,flag=0;
}node[maxn],temp;
int main(void){
    int s1,s2,N,i,address,next;
    char data;
    scanf("%d%d%d",&s1,&s2,&N);
    for(i=0;i<N;i++){
        scanf("%d %c %d",&address,&data,&next);
        node[address].data=data;
        node[address].next=next;
    }
    while(s1!=-1){
        node[s1].flag=1;
        s1=node[s1].next;
    }
    while(s2!=-1){
        if(node[s2].flag==1){
            printf("%05d",s2);
            return 0;
        }
        s2=node[s2].next;
    }
    printf("-1");
}

【解决方案2】

#include<stdio.h>
#include<string.h>
//A1032 监测点0 1:11分 3：8分 4：1分
//19-2-19 少用do while
struct node{
    int address,next;
    char data;
};
typedef struct node NODE;
NODE node1[100005],temp;
int hashtable[100005]={0},addresstable[100005]={0};
char str1[1000005],str2[1000005];
void reverse(char str[]){
    int i,j,len;
    char ch;
    len=strlen(str);
    for(i=0,j=len-1;i<=j;i++,j--){
        ch=str[i];
        str[i]=str[j];
        str[j]=ch;
    }
}
int main(){
    int start1,start2,end1,end2,i,N,same,flag=0;
    scanf("%d%d%d",&start1,&start2,&N);
    for(i=0;i<N;i++){
        scanf("%d %c%d",&node1[i].address,&node1[i].data,&node1[i].next);
        addresstable[node1[i].address]=i;
    }
    i=0;
    while(start1!=-1){
        temp=node1[addresstable[start1]];
        hashtable[temp.address]=1;
        str1[i++]=temp.data;
        start1=temp.next;
    }
    end1=temp.address;
    str1[i]='\0';
    i=0;
    while(start2!=-1){
        temp=node1[addresstable[start2]];
        str2[i++]=temp.data;
        start2=temp.next;
        if(hashtable[temp.address]==1){
            printf("%05d",temp.address);
            return 0;
        }
    }
    end2=temp.address;
    str2[i]='\0';
    reverse(str1);
    reverse(str2);
    if(str1[0]!=str2[0]||end1!=end2){
        printf("-1");
        return 0;
    }
    for(i=0;str1[i]==str2[i];i++);
    i--;
    char ch=str1[i];
    for(i=0;i<N;i++){
        if(node1[i].data==ch)
            break;
    }
    printf("%05d",node1[i].address);
    return 0;
}