题目1468：Sharing

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).

输入：

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

where Address 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.

输出：

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.

样例输入：

11111 22222 967890 i 0000200010 a 1234500003 g -112345 D 6789000002 n 0000322222 B 2345611111 L 0000123456 e 6789000001 o 0001000001 00002 400001 a 1000110001 s -100002 a 1000210002 t -1

样例输出：

67890-1

来源：

2012年浙江大学计算机及软件工程研究生机试真题

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int ft[111111]={0};int findlen(int n){    int t=n,len=1;    while(ft[t]>0)    {        t=ft[t];        len++;    }    return len;}int findsuffic(int a1,int a2,int tmp){    while(tmp--)    {        a1=ft[a1];    }    while(a1!=a2)    {        a1=ft[a1];        a2=ft[a2];    }    if(a1>0)        printf("%05d\n",a1);    else        printf("-1\n");}int main(){    int a,b,n,i,j;    while(std::cin>>a>>b>>n)    {        while(n--)        {            char c;            int t1,t2;            scanf("%d %c %d",&t1,&c,&t2);            ft[t1]=t2;        }        int len1,len2;        len1=findlen(a);        len2=findlen(b);            if(len1>len2)        {            findsuffic(a,b,len1-len2);        }        else        {            findsuffic(b,a,len2-len1);//这里不小心写成len1-len2了，然后无限TLE,找了好久才找出错来，下次一定要认真！        }    }    return 0;}