poj1269 Intersecting Lines

传送门

本来是奔着做线段求交去的qwq

结果是个直线求交...

直接用课本知识求个一般式然后搞就行

(话说这个代码应该加一句a1==0||a2==0时把a变成b)

但是poj数据水就过了

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
typedef double D;
#define eps 1e-8
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
//head
#define R(x) scanf("%lf",&x)
D x1,x2,x3,x4;
D y1,y2,y3,y4;
D a1,b1,c1;
D a2,b2,c2;
D d,ansx,ansy;
void solve() {
    R(x1),R(y1),R(x2),R(y2),R(x3),R(y3),R(x4),R(y4);
    a1 = x1-x2,b1 = y2-y1,c1 = x1*y2-x2*y1;
    a2 = x3-x4,b2 = y4-y3,c2 = x3*y4-x4*y3;
    d = a1*b2-a2*b1;
    if(d == 0) return void(c1/a1 == c2/a2 ? puts("LINE") : puts("NONE"));
    D ansy = (b2*c1-b1*c2) / d;
    D ansx = (a1*c2-a2*c1) / d;
    printf("POINT %.2lf %.2lf
",ansx,ansy);
}

int main() {
    int T = read();
    puts("INTERSECTING LINES OUTPUT");
    while(T--) solve();
    puts("END OF OUTPUT");
    return 0;
}