B. Dinner with Emma
题目连接：
http://www.codeforces.com/contest/616/problem/A
Description
Jack decides to invite Emma out for a dinner. Jack is a modest student, he doesn't want to go to an expensive restaurant. Emma is a girl with high taste, she prefers elite places.
Munhattan consists of n streets and m avenues. There is exactly one restaurant on the intersection of each street and avenue. The streets are numbered with integers from 1 to n and the avenues are numbered with integers from 1 to m. The cost of dinner in the restaurant at the intersection of the i-th street and the j-th avenue is cij.
Jack and Emma decide to choose the restaurant in the following way. Firstly Emma chooses the street to dinner and then Jack chooses the avenue. Emma and Jack makes their choice optimally: Emma wants to maximize the cost of the dinner, Jack wants to minimize it. Emma takes into account that Jack wants to minimize the cost of the dinner. Find the cost of the dinner for the couple in love.
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 100) — the number of streets and avenues in Munhattan.
Each of the next n lines contains m integers cij (1 ≤ cij ≤ 109) — the cost of the dinner in the restaurant on the intersection of the i-th street and the j-th avenue.
Output
Print the only integer a — the cost of the dinner for Jack and Emma.
Sample Input
3 4
4 1 3 5
2 2 2 2
5 4 5 1
Sample Output
2
Hint
题意
给你n行m列，A负责选行，B负责选列，A先选
A希望选择出来的数尽量大，B希望选择出来的数尽量小
问最后答案应该是多少
题解：
输出比较每一行的最小值，然后输出最大的就好了
因为A先选，所以肯定会选择某一行中，最小值最大的那个
B肯定选择这一行的最小值
代码
#include<bits/stdc++.h>
using namespace std;
int Ans = 0;
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
{
int T = 1e9;
for(int j=0;j<m;j++)
{
int x;scanf("%d",&x);
T = min(x,T);
}
Ans = max(T,Ans);
}
printf("%d
",Ans);
}