1032. Sharing (25)
来源:互联网 发布:js弹出div模态窗口 编辑:程序博客网 时间:2024/06/05 10:58
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 967890 i 0000200010 a 1234500003 g -112345 D 6789000002 n 0000322222 B 2345611111 L 0000123456 e 6789000001 o 00010Sample Output 1:
67890Sample Input 2:
00001 00002 400001 a 1000110001 s -100002 a 1000210002 t -1Sample Output 2:
-1
#include<cstdio>#include<cstdlib>#define Maxn 100001typedef struct{int address;char key;int next;}Node;Node input[Maxn];char test[Maxn];int main(void){int N,address1,address2;scanf("%5d %5d %d",&address1,&address2,&N);int i;int inaddr,innext;char inkey;for(i=0;i<N;++i){scanf("%5d %c %d",&inaddr,&inkey,&innext);input[inaddr].address = inaddr;input[inaddr].key = inkey;input[inaddr].next = innext;}int header1 = address1;while(header1 != -1){test[header1] = input[header1].key;header1 = input[header1].next;}bool havecommon = false;int header2 = address2;while(header2 != -1){if(input[header2].key == test[header2]){havecommon = true;break;}header2 = input[header2].next;}if(havecommon){printf("%05d",header2);}else{printf("-1");}system("pause");return 0;}
- 1032. Sharing (25)
- 1032. Sharing (25)-PAT
- 【PAT】1032. Sharing (25)
- 1032. Sharing (25)
- 1032. Sharing (25)
- PAT 1032. Sharing (25)
- PAT 1032. Sharing (25)
- PAT 1032. Sharing (25)
- PAT:1032. Sharing (25)
- 1032. Sharing (25)
- PAT 1032. Sharing (25)
- 1032. Sharing (25)
- 1032. Sharing (25)
- 1032. Sharing (25)
- 1032. Sharing (25)
- 1032. Sharing (25)
- 1032. Sharing (25)
- pat 1032. Sharing (25)
- Linux设备驱动程序简介
- 魔方阵
- C 专家编程的一些小知识点
- android 图片缩放手势
- android 转场之transition
- 1032. Sharing (25)
- JSON 和 XML 优缺点的比较
- Linux 设备驱动模型
- VC++中字符串编码的转换
- Linux Shell根据进程名杀死进程
- JS转换成带两位小数的数值
- HTML5中使用不同type的属性将会呈现不同的键盘样式
- ELaticSearch——PlainElastic.net.dll 中的JsonNetSerializer
- URL编码解码原理及演示(Java演示)