uva 1610——Party Games

来源：互联网发布：淘宝买恒压阀警察找我编辑：程序博客网时间：2024/06/05 15:29

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=51171

题意：给你n个串的集合D，然后求一个长度最短的串s，使得使得s大于等于D中一半的串，又同时小于另一半串。

思路：直接暴力。先对n个串排序，然后选择第n/2-1个串，从第一个字母开始找，在后面添加26个字母枚举，直至找到目标串。

code：

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <sstream>#include <string>#include <vector>#include <list>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef long double ld;const int INF=0x3fffffff;const int inf=-INF;const int N=1000000;const int M=2005;const int mod=1000000007;const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))#define cpy(x,a) memcpy(x,a,sizeof(a))#define fr(i,s,n) for (int i=s;i<=n;i++)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lrt  rt<<1#define rrt  rt<<1|1#define middle int m=(r+l)>>1#define lowbit(x) (x&-x)#define pii pair<int,int>#define mk make_pair#define IN freopen("in.txt","r",stdin);#define OUT freopen("out.txt","w",stdout);string s[M];int n;void sol(){    sort(s,s+n);    //fr(i,0,n-1) cout<<s[i]<<endl;    string s1=s[n/2-1],s2=s[n/2],t="",p="";    for (int i=0;;++i)    {        for (char c='A';c<='Z';++c)        {                t=p+c;            if(t>=s1&&t<s2){cout<<t<<endl;return;}        }        p+=s1[i];    }}int main(){    while (~scanf("%d",&n)&&n)    {        fr(i,0,n-1) cin>>s[i];        sol();    }}

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