CodeForces 1288D Minimax Problem （二分）

Description：

You are given $n$ arrays $a_{1}, a_{2}, ..., a_{n}$; each array consists of exactly m integers. We denote the $y-th$ element of the $x-th$ array as $a_{x,y}$.

You have to choose two arrays KaTeX parse error: Expected '}', got 'EOF' at end of input: a_{i and KaTeX parse error: Expected '}', got 'EOF' at end of input: a_{j ( $1≤i,j≤n$, it is possible that $i=j$). After that, you will obtain $a$ new array $b$ consisting of $m$ integers, such that for every $k∈[1,m] ，b_{k}=max(a_{i,k},a_{j,k})$.

Your goal is to choose $i$ and $j$ so that the value of $min_{k=1}^{m}b_{k}$ is maximum possible.

Input

The first line contains two integers $n$ and $m$ ( $1≤n≤3⋅10^5, 1≤m≤8$) — the number of arrays and the number of elements in each array, respectively.

Then $n$ lines follow, the $x-th$ line contains the array $a_{x}$ represented by m integers $a_{x,1}, a_{x,2}, ..., a_{x,m} (0≤a_{x,y}≤10^9)$.

Output

Print two integers $i$ and $j$ ( $1≤i,j≤n$, it is possible that $i=j$) — the indices of the two arrays you have to choose so that the value of $min_{k=1}^{m}b_{k}$ is maximum possible. If there are multiple answers, print any of them.

Example

input

6 5
5 0 3 1 2
1 8 9 1 3
1 2 3 4 5
9 1 0 3 7
2 3 0 6 3
6 4 1 7 0

output

1 5

题意：

给出 $n$ 个长度为 $m$ 的数组，在两个数组中选取位置一样中大的那个元素，然后求新组成的数组中的元素最小值，然后求全部最小值中的最小值，输出是由哪两个数组组合成的。
二分答案，二进制优化，具体解释看代码吧。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
    if (a >= mod)
        a = a % mod + mod;
    ll ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = ans * a;
            if (ans >= mod)
                ans = ans % mod + mod;
        }
        a *= a;
        if (a >= mod)
            a = a % mod + mod;
        b >>= 1;
    }
    return ans;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}

///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //算出它们累乘的结果
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //计算逆元
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}

int n, m;
int val[(1 << 8) + 10], pos[(1 << 8) + 10];
int a[300010][10];
int l = inf, r = -1;
int ans1, ans2;
bool judge(int x)
{
    rep(i, 1, n)
    {
        int t = 0;
        rep(j, 0, m - 1)
        {
            if (a[i][j + 1] < x)
                t |= (1 << j); //对t在二进制下的第j位赋值为1，t是第二个序列选择的情况，
                //第一个序列选择的都是比x大的数据，这样保证x是新序列最小的
        }
        if (val[t] >= x)
        {
            ans1 = i;
            ans2 = pos[t];
            if (t == 0)
                ans2 = i;
            return true;
        }
    }
    return false;
}
int main()
{
    mem(val, 0);
    mem(pos, 0);
    val[0] = inf;
    sd(n);
    sd(m);
    rep(i, 1, n)
    {
        rep(j, 1, m)
        {
            sd(a[i][j]);
            l = min(l, a[i][j]); //全部的最小值
            r = max(r, a[i][j]); //全部的最大值
        }
    }
    rep(i, 1, n)
    {
        rep(j, 1, (1 << m) - 1)
        { //j就是在每一个序列里一个有2^m种选取情况，然后对于每种情况求最小值
            int temp = inf;
            rep(k, 0, m - 1)
            {
                if (j & (1 << k)) //j的第k位是不是1
                    temp = min(temp, a[i][k + 1]);
            }
            if (temp > val[j])
            {
                val[j] = temp;
                pos[j] = i;
            }
        }
    }
    int mid, ans = -1, t = l;
    while (l <= r)
    {
        mid = (l + r) >> 1;
        if (judge(mid))
            l = mid + 1, ans = mid;
        else
            r = mid - 1;
    }
    if (ans == -1) //处理边界
        ans = t;
    judge(ans);
    pdd(ans1, ans2);
    return 0;
}