PAT 1032 Sharing

1032 Sharing (25 分)

 

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 (≤), 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 −.

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

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MAXN 100005

int vis[MAXN] = {0};


int main(){
    int a,b,n;
    cin >> a >> b >> n;
    map<int,int> mp;
    for(int i=0;i < n;i++){
        int x,y;
        char t;
        scanf("%d %c %d",&x,&t,&y);
//cin >> x >> t >> y;
        mp[x] = y;
    }

    while(a!=-1){
        vis[a] = 1;
        a = mp[a];
    }

    while(1){
        if(b==-1){printf("-1");break;}
        if(vis[b]){printf("%05d",b);break;}
        b = mp[b];
    }
//    int temp = b;
//    while(1){
//        if(a==-1){printf("-1");break;}
//        b = temp;
//        while(1){
//            if(b==-1){ break;}
//            if(a == b){printf("%05d",a);return 0;}
//            b = mp[b];
//        }
//        a = mp[a];
//    }


    return 0;
}

一开始写了O(N^2)的暴力，后来改了遍历第一条链表，将访问过的结点的flag都标记为true，当遍历第二条结点的时候，如果遇到了true的结点就输出并结束程序，没有遇到就输出-1

然后输入又被卡，cin过不了（以后发现输入大于10^5就用scanf吧）char --> %c  输入要空格和printf一样，真是麻烦死