URAL 2067 Friends and Berries (推理，数学)

时间：2023-02-02 20:33:25

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;

}

　　

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