CodeForces 441A Valera and Antique Items
链接：http://codeforces.com/problemset/problem/441/A
Valera and Antique Items
Valera is a collector. Once he wanted to expand his collection with exactly one antique item.
Valera knows n sellers of antiques, the i-th of them auctioned ki items. Currently the auction price of the j-th object of the i-th seller is sij. Valera gets on well with each of the n sellers. He is perfectly sure that if he outbids the current price of one of the items in the auction (in other words, offers the seller the money that is strictly greater than the current price of the item at the auction), the seller of the object will immediately sign a contract with him.
Unfortunately, Valera has only v units of money. Help him to determine which of the n sellers he can make a deal with.
Input
The first line contains two space-separated integers n, v (1 ≤ n ≤ 50; 104 ≤ v ≤ 106) — the number of sellers and the units of money the Valera has.
Then n lines follow. The i-th line first contains integer ki (1 ≤ ki ≤ 50) the number of items of the i-th seller. Then go ki space-separated integers si1, si2, ..., siki (104 ≤ sij ≤ 106) — the current prices of the items of the i-th seller.
Output
In the first line, print integer p — the number of sellers with who Valera can make a deal.
In the second line print p space-separated integers q1, q2, ..., qp (1 ≤ qi ≤ n) — the numbers of the sellers with who Valera can make a deal. Print the numbers of the sellers in the increasing order.
Sample test(s)
3 50000 1 40000 2 20000 60000 3 10000 70000 190000
3 1 2 3
3 50000 1 50000 3 100000 120000 110000 3 120000 110000 120000
0
Note
In the first sample Valera can bargain with each of the sellers. He can outbid the following items: a 40000 item from the first seller, a 20000 item from the second seller, and a 10000 item from the third seller.
In the second sample Valera can not make a deal with any of the sellers, as the prices of all items in the auction too big for him.
大意——有n个古董卖家，第i个卖家有ki个古董要拍卖。当前知道第i个卖家的第j件古董的拍卖价格为sij，一个人和n个卖家的关系都比较好，他非常确定只要他出价高于第i个卖家的古董之一的当前价格，第i个卖家就会和他达成协议。但不幸的是，当前他仅仅只有v单位的钱了，问：他当前能和几个卖家达成协议，分别是哪几个(卖家按照标号呈上升序列给出)？
思路——简单模拟一下即可。将当前钱数与第i个卖家的所有古董拍卖价格进行比较，一旦发现当前钱数大于古董拍卖价格的情况，就记录可以和该卖家达成协议了，然后用一个数组记录卖家信息，数组下标表示卖家个数，数组元素表示卖家标号。最后和所有卖家古董拍卖价格都比较完了，直接输出最大数组下标，然后按顺序输出记录卖家标号的数组元素即可。
复杂度分析——时间复杂度:O(nk),空间复杂度:O(cnt)
附上AC代码：
#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
typedef unsigned int li;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
short n, k;
int extra, value;
int main()
{
ios::sync_with_stdio(false);
while (~scanf("%hd%d", &n, &extra))
{
short num[55], cnt=0;
for (short i=0; i<n; i++)
{
scanf("%hd", &k);
bool flag = false;
while (k--)
{
scanf("%d", &value);
if (extra > value)
flag = true;
}
if (flag)
num[cnt++] = i+1;
}
printf("%hd\n", cnt);
for (short i=0; i<cnt; i++)
printf("%d ", num[i]);
printf("\n");
}
return 0;
}