题意：给定 n 个人，每个人两个值s, r，要满足，p(v, u) = sqrt((sv − su)^2 + (rv − ru)^2)， p(v,u,w) = (p(v,u) + p(v,w) + p(u,w)) / 2
要求找出p(v, u) ≥ p(v,u,w) 的对数，其中w是除u，v外，任意的人。
析：化简一下这个表达式，这很明显是两边之和小于等于第三边，好像挺熟悉啊，对，三角形是两边之和大于第三边，现在是小于，不可能，只能是等于，要想等于，
那么只是共线了，并且w是任何人，所以这些点全部都要共线。并且两端的人就是答案，并且只有一对。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e5 + 5;
const int mod = 1e8;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct node{
LL x, y;
int id;
node() { }
node(LL xx, LL yy) : x(xx), y(yy){ }
bool operator < (const node &p) const{
return x < p.x || (x == p.x && y < p.y);
}
};
node a[maxn]; int main(){
while(scanf("%d", &n) == 1){
LL x, y;
for(int i = 0; i < n; ++i){
scanf("%I64d %I64d", &x, &y);
a[i] = node(x, y);
a[i].id = i+1;
}
bool ok = true;
for(int i = 2; i < n; ++i)
if((a[i-1].x-a[i].x)*(a[i-2].y-a[i].y) != (a[i-1].y-a[i].y)*(a[i-2].x-a[i].x)){ ok = false; break;} if(!ok){ printf("0\n"); continue; }
sort(a, a+n);
printf("1\n%d %d\n", a[0].id, a[n-1].id);
}
return 0;
}