UVA 1610 Party Games（聚会游戏）（构造）

题意：输入一个n(2<=n<=1000，n是偶数)个字符串的集合D，找一个长度最短的字符串S(不一定在D中出现)，使得D中恰好一半串小于等于S，另一半串大于S。如果有多解，输出字典序最小的解。

分析：找到最中间的两个串，直接按位构造。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 1e3 + 10;
const int MAXT = 10000 + 10;
using namespace std;
string s[MAXN];
string solve(int n){
    int mid = n / 2;
    int len = s[mid - 1].size();
    string ans = "";
    ans += 'A';
    int i = 0;
    while(i < len){
        while(ans[i] < 'Z' && ans < s[mid - 1]) ++ans[i];//只要i不是len-1且s[mid - 1][i]不是Z，得到的ans[i]都会比s[mid - 1]恰好大1,而如果i等于len-1，则得到的ans与s[mid-1]正好相等
        if(ans >= s[mid - 1] && ans < s[mid]) return ans;
        if(s[mid - 1][i] != ans[i]) --ans[i];//如果s[mid - 1][i]不是Z，需要减1
        ans += 'A';
        ++i;
    }
}
int main(){
    int n;
    while(scanf("%d", &n) == 1){
        if(!n) return 0;
        for(int i = 0; i < n; ++i) cin >> s[i];
        sort(s, s + n);
        printf("%s\n", solve(n).c_str());
    }
    return 0;
}