/
本题收获：
1. map 的插入
格式如下：
typedef map mymap;
mymap m;
m.insert(pair(vs,100)); //pair可简写为 make_pair
有关博客:
2. vector的排序和去重
去重之前需要先排序！因为unique和erase一起用的作用，仅仅是删除相邻的重复元素，所以必须先进行排序
核心代码：
vectorv;
sort(v.begin(), v.end());
v.erase(unique(v.begin(), v.end()), v.end());
------------------------------------------
同时，之前学过的坐标离散化，也用到了vector的排序和去重
/
A. Banana
/
题目链接：
2017 ACM-ICPC 亚洲区（乌鲁木齐赛区） A. Banana
Bananas are the favoured food of monkeys.
In the forest, there is a Banana Company that provides bananas from different places.
The company has two lists.
The first list records the types of bananas preferred by different monkeys, and the second one records the types of bananas from different places.
Now, the supplier wants to know, whether a monkey can accept at least one type of bananas from a place.
Remenber that, there could be more than one types of bananas from a place, and there also could be more than one types of bananas of a monkey's preference.
Input Format
The first line contains an integer TT, indicating that there are TT test cases.
For each test case, the first line contains two integers NN and MM, representing the length of the first and the second lists respectively.
In the each line of following NN lines, two positive integers i, ji,j indicate that the ii-th monkey favours the jj-th type of banana.
In the each line of following MM lines, two positive integers j, kj,k indicate that the jj-th type of banana could be find in the kk-th place.
All integers of the input are less than or equal to 5050.
Output Format
For //代码效果参考：http://www.lyjsj.net.cn/wz/art_22724.html
each test case, output all the pairs x, yx,y that the xx-the monkey can accept at least one type of bananas from the yy-th place.These pairs should be outputted as ascending order. That is say that a pair of x, yx,y which owns a smaller xx should be output first.
If two pairs own the same xx, output the one who has a smaller yy first.
And there should be an empty line after each test case.
样例输入
1
6 4
1 1
1 2
2 1
2 3
3 3
4 1
1 1
1 3
2 2
3 3
样例输出
1 1
1 2
1 3
2 1
2 3
3 3
4 1
4 3
/
#include
#include
#include
#include
using namespace std;
map banana; // key: number of monkeys, value: type of bananas;
map location; // key: type of bananas; value: location of bananas;
int main()
{
int t, n, m, x, y;
cin ] t;
while (t--)
{
banana.clear();
location.clear();
cin ] n ] m;
for (int i = 0; i < n; i++)
{
cin ] x ] y;
if (banana.count(x) == 0)
{
vector tp;
tp.push_back(y);
banana.insert(make_pair(x, tp));
}
else
{
vector tp = banana【x】;
tp.push_back(y);
banana【x】 = tp;
}
}
for (int i = 0; i < m; i++)
{
cin ] x ] y;
if (location.count(x) == 0)
{
vector tp;
tp.push_back(y);
location.insert(make_pair(x, tp));
}
else
{
vector tp = location【x】;
tp.push_back(y);
location【x】 = tp;
}
}
for (int i = 1; i <= n; i++)
{
vectorv;
for (int j = 0; j < banana【i】.size(); j++)
{
int tp = banana【i】【j】;
if (location.count(tp) != 0)
{
vector a = location【tp】;
for (int ii = 0; ii < a.size(); ii++)
v.push_back(a【ii】);
}
}
sort(v.begin(), v.end());
v.erase(unique(v.begin(), v.end()), v.end());
for (int k = 0; k < v.size(); k++)
cout [ i [ " " [ v【k】 [ endl;
}
cout [ endl;
}
}
