PAT 1032. Sharing (25)（求俩个字符串的交点）

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1032. Sharing (25)

时间限制
100 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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 (<= 105), 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.

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

解题思路

• 1.长的那个减掉长的个数，然后在比较，找到地址相同了的即可。

AC代码

#include<iostream>#include<cstdio>using namespace std;char value[99999];int next1[99999];int main(){    //输入    int f,s,n;    scanf("%d %d %d",&f,&s,&n);    int where,to;char a;    for (int i= 0; i < n; ++i) {        scanf("%d %c %d",&where,&a,&to);        value[where] = a;        next1[where] = to;    }    //分别计算俩个链表的长度la和lb    int la = 0,lb = 0;    int now = f;    while (now != -1) {        la++;        now = next1[now];    }    now = s;    while (now != -1) {        lb++;        now =next1[now];    }   // cout << la << lb;    //长的先走    int nowa = f;    int nowb =s;    if (la>lb) {        int d = la - lb;        while (d--) {            nowa = next1[nowa];        }    }else {        int d = lb - la;        while (d--) {            nowb =next1[nowb];        }    }    //找到相等了的节点即可    while (nowa !=nowb) {        nowa = next1[nowa];        nowb = next1[nowb];    }    if (nowa == -1) {        printf("%d\n",-1);    }else {        printf("%05d\n",nowa);    }}