Codeforces-1019D：Large Triangle（暴力+Hash）

D. Large Triangle
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

There is a strange peculiarity: if you connect the cities of Rostov, Taganrog and Shakhty, peculiarly, you get a triangle
«Unbelievable But True»
Students from many different parts of Russia and abroad come to Summer Informatics School. You marked the hometowns of the SIS participants on a map.

Now you decided to prepare an interesting infographic based on this map. The first thing you chose to do is to find three cities on this map, such that they form a triangle with area S.

Input
The first line of input contains two integers $n$ and $S$ $(3≤n≤2000, 1≤S≤2⋅1018)$ — the number of cities on the map and the area of the triangle to be found.

The next n lines contain descriptions of the cities, one per line. Each city is described by its integer coordinates $x_i, y_i (−10^9≤x_i,y_i≤10^9)$.

It is guaranteed that all cities are located at distinct points. It is also guaranteed that no three cities lie on the same line.

Output
If the solution doesn’t exist — print «No».

Otherwise, print «Yes», followed by three pairs of coordinates (x,y) — the locations of the three cities, which form the triangle of area S.

Examples
input
3 7
0 0
3 0
0 4
output
No
input
4 3
0 0
2 0
1 2
1 3
output
Yes
0 0
1 3
2 0

思路：看CF大神代码，收获颇多。暴力枚举三点，判读是否存在满足条件的三点。
但是在计算中不要用long long防溢出，直接让它自然溢出，这就相当于hash了，同时int型的运算也比long long 型的运算更快。如果发现满足条件的三个点，然后再用long long型的判断一下Hash是否冲突即可。

#include<bits/stdc++.h>
using namespace std;
const int MAX=2e3+10;
const int MOD=1e9+7;
const int INF=1e9+7;
const double PI=acos(-1.0);
typedef long long ll;
inline ll read()//加的快读，貌似没一点儿用。。
{
    ll f=1,x=0;char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    return f*x;
}
int x[2000],y[2000];
int main()
{
    int n=read();
    ll sum=read()*2;
    int S=sum;
    for(int i=0;i<n;i++)x[i]=read(),y[i]=read();
    for(int i=0;i<n;i++)
    {
        for(int j=i+1;j<n;j++)x[j]-=x[i],y[j]-=y[i];
        for(int j=i+1;j<n;j++)
        for(int k=j+1;k<n;k++)
        {
            int ss=x[k]*y[j]-x[j]*y[k];//自然溢出
            if((S==ss||S==-ss)&&(sum==abs(1ll*x[k]*y[j]-1ll*x[j]*y[k])))//判断是否满足条件
            {
                puts("Yes");
                printf("%d %d\n",x[i],y[i]);
                printf("%d %d\n",x[i]+x[j],y[i]+y[j]);
                printf("%d %d\n",x[i]+x[k],y[i]+y[k]);
                return 0;
            }
        }
        for(int j=i+1;j<n;j++)x[j]+=x[i],y[j]+=y[i];
    }
    puts("No");
    return 0;
}