九度oj 题目1468：Sharing 【ZJU2012考研机试题2】

题目1468：Sharing

时间限制：1 秒

内存限制：128 兆

特殊判题：否

提交：1744

解决：338

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

涉及到地址和值的问题，就是用Hash存储思想优化。

/**************************************************************    Problem: 1468    Language: C++    Result: Accepted    Time:190 ms    Memory:1800 kb****************************************************************/#include <stdio.h>#include<string.h>const int MAX=100010;int hash[MAX];int mark[MAX];int main(){    int a,b,c,d,n,i;    char e;    //freopen("G:\\in.txt","r",stdin);    while(scanf("%d%d%d",&a,&b,&n)!=EOF){        int start1=a;        int start2=b;        for(i=0;i<n;i++){            scanf("%d %c %d",&c,&e,&d);            hash[c]=d;        }        for(i=0;i<MAX;i++)            mark[i]=0;        int id1=start1,id2=start2;        while(id1!=-1){            mark[id1]=1;            id1=hash[id1];        }        while(id2!=-1){            if(mark[id2]==1){                printf("%05d\n",id2);                break;            }            else                id2=hash[id2];        }        if(id2==-1)            printf("-1\n");    }      return 0;}