九度oj 1468 Sharing 2012年浙江大学计算机及软件工程研究生机试真题

题目1468：Sharing

时间限制：1 秒

内存限制：128 兆

特殊判题：否

提交：2687

解决：550

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 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
00001 00002 4
00001 a 10001
10001 s -1
00002 a 10002
10002 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

 

 

分析：

如果有公共节点，则第一个公共节点必同时作为两个节点的后驱，且该两个节点分别属于a,b字符串。因此，只要统计后继节点的出现次数，如果有出现两次的节点，则该节点即为所求。

另外，还要注意最后的输出格式问题。

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int ex[100005];
int main(){
    int f1,f2,n;
    while(cin>>f1>>f2>>n){
        int get=-1;
        int i;
        int f;
        char c;
        memset(ex,0,sizeof(ex));
        for(i=0;i<n;i++){
            cin>>f;
            cin>>c;
            cin>>f;
            if(f==-1||get!=-1){
                continue;
            }
            ex[f]+=1;
            if(ex[f]==2){
                get=f;
            }
        }
        if(get==-1){
            cout<<get<<endl;
        }
        else{
            printf("%05d\n",get);//格式问题注意！！ 
        }
    }
    return 0;
}

 

网上找的做法：
思想可以借鉴：

1.a=node[a];
2.出现了两次的点就是所求

#include<iostream>
#include<vector>
#include<cstring>
#include<cstdio>
using namespace std;
const int MAXN=100000;
int flag[MAXN];
int node[MAXN];
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
#endif
    int ad1,ad2,N;
    scanf("%d%d%d",&ad1,&ad2,&N);
    int a(0),b(0);
    memset(node,0,sizeof(node));
    memset(flag,0,sizeof(flag));
    char c;
    while(N--)
    {
       scanf("%d %c %d",&a,&c,&b);
       node[a]=b;
    }
    a=ad1;
    while(a!=-1)
    {
        flag[a]=1;
        a=node[a];
    }
    b=ad2;
    //while(b!=-1&&!flag[b])
    while(!flag[b]&&b!=-1)
    {
        //flag[b]=1;
        b=node[b];
    }
    if(b!=-1)
        printf("%05d\n",b);//格式问题呀！！！一定要注意！！！
    else 
        printf("-1\n");
    return 0;
}