POJ 2737 Swamp Things （斜率判断多点共线）

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Swamp Things

题意: 判断最多的共线的点数

思路: 把所有点对的斜率都存下来然后排序看有多少相同的相同的说明共线注意判断斜率不存在的情况

参考code:

map TLE了慎用STL

////  Created by TaoSama on 2015-06-12//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int N = 1e5 + 10;const double EPS = 1e-10;int n, x[1005], y[1005];double k[1005];int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int kase = 0;    while(scanf("%d", &n) == 1 && n) {        for(int i = 1; i <= n; ++i) scanf("%d%d", x + i, y + i);        int ans = 0;        for(int i = 1; i <= n; ++i) {            int cnt = 0;            for(int j = i + 1; j <= n; ++j) {                if(x[i] == x[j]) k[cnt++] = INF;                else k[cnt++] = (y[i] - y[j]) * 1.0 / (x[i] - x[j]);            }            sort(k, k + cnt);            for(int j = 0; j < cnt; ++j) {                int ct = 1;                while(j + 1 < cnt && abs(k[j] - k[j + 1]) < EPS)                    ++j, ++ct;                ans = max(ct, ans);            }        }        ans += 1;        if(ans < 4) ans = 0;        printf("Photo %d: %d points eliminated\n", ++kase, ans);    }    return 0;}/*int n, x[1005], y[1005];int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int kase = 0;    while(scanf("%d", &n) == 1 && n) {        for(int i = 1; i <= n; ++i) scanf("%d%d", x + i, y + i);        int ans = 0;        for(int i = 1; i <= n; ++i) {            int Max = 0;            map<double, int> slope;            for(int j = i + 1; j <= n; ++j) {                double t;                if(x[i] == x[j]) t = INF;                else t = (y[i] - y[j]) * 1.0 / (x[i] - x[j]);                //cout << t << endl;                if(!slope.count(t)) {                    slope[t] = 2;                    Max = max(Max, 2);                } else {                    slope[t]++;                    Max = max(Max, slope[t]);                }            }            ans = max(ans, Max);        }        if(ans < 4) ans = 0;        printf("Photo %d: %d points eliminated\n", ++kase, ans);    }    return 0;}*/

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