SCU4490 Lisp em(string和map容器的应用)

题目：

Time Limit: 1000 MS    Memory Limit: 131072 K     
Description
There are two lists and they may be intersected with each other.You must judge if they are intersected and find the first node they have in common.
Input
The first line is the number of cases.Each case consists of two lists.Each list consists of multiple nodes.Each node is represented by its address in memory.So, a list may look like this:    0x233333    0x123456    0xabcdef1   0xffffff    nilnil is the end of the list.All lists have no cycle.The length of list is not larger than 4e5.List can't be empty.Warn: huge input.
Output
"No" if they are not intersected."Yes Address", if they are intersected, print "Yes" and the address of their first common node.
Sample Input
20x233333    0x123456    0xabcdef1   0xffffff    nil0x999999    0x332232    0xffffff    nil0x233333    0x123456    0xabcdef1   0xffffff    nil0x987654    0xafafaf    0xfffcccc   0xaaface    nil
Sample Output
Yes 0xffffffNo
Author
qw4990

思路;

这就没什么思路了，一看代码肯定懂了，注意细节就好，存一存代码~

代码：

#include <stdio.h>#include <stdlib.h>#include <math.h>#include <iostream>#include <cstring>#include <vector>#include <map>#include <string>#define N 500#define LL long long#define mem(a,b) memset(a,b,sizeof(a))using namespace std;string s;map<string,int>mp;vector<string>a;int main(){    int t;    scanf("%d",&t);    while(t--)    {        s.clear();        a.clear();        mp.clear();        while(cin>>s&&s!="nil")            mp[s]=1;        while(cin>>s&&s!="nil")            if(mp[s]==1)                a.push_back(s);        if(a.size()==0)            printf("No\n");        else            cout<<"Yes"<<" "<<a[0]<<endl;    }    return 0;}