# 什么是羁绊## 题目背景JJY是一名资深炉石玩家自从国服停运JJY就转战外服并凭借他的羁绊用一套愚人法在天梯叱咤风云。## 题目描述目前JJY的套牌中有 $n$ 张卡牌JJY对每张卡牌从法力消耗、法术效果和稀有度三个方面进行了评分第 $i$ 张卡牌的评分是 $a_i$$b_i$$c_i$。JJY希望改进自己的套牌。众所周知JJY是富哥他可以升级任意张卡牌第 $i$ 张卡牌升级的评分是 $a_i

#include<iostream>
#include<algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
int a[maxn], b[maxn], c[maxn], ap[maxn], bp[maxn], cp[maxn];
int main(){
    int n;
    cin >> n;
    int maxa = -INF, mina = INF, maxb = -INF, minb = INF, maxc = -INF, minc = INF;
    for(int i = 1; i <= n; i++){
        cin >> a[i] >> b[i] >> c[i] >> ap[i] >> bp[i] >> cp[i];
        maxa = max(maxa, a[i]);
        mina = min(mina, a[i]);
        maxb = max(maxb, b[i]);
        minb = min(minb, b[i]);
        maxc = max(maxc, c[i]);
        minc = min(minc, c[i]);
    }
    int ans = max(max(maxa - mina, maxb - minb), maxc - minc);
    for(int i = 1; i <= n; i++){
        int tmpa = max(maxa, ap[i]);
        int tmpb = max(maxb, bp[i]);
        int tmpc = max(maxc, cp[i]);
        ans = min(ans, max(max(tmpa - mina, tmpb - minb), tmpc - minc));
    }
    cout << ans << endl;
    return 0;
}
``