注释上都有解析了，就不写了吧，去重的问题就用set解决，并且呢第i个线段最多和其他线段产生i-1个交点,n^2logn。
#include <cmath> #include <cstdio> #include <cstdlib> #include <cassert> #include <cstring> #include <set> #include <map> #include <list> #include <queue> #include <string> #include <iostream> #include <algorithm> #include <functional> #include <stack> using namespace std; typedef long long ll; #define T int t_;Read(t_);while(t_--) #define dight(chr) (chr>='0'&&chr<='9') #define alpha(chr) (chr>='a'&&chr<='z') #define INF (0x3f3f3f3f) #define maxn (300005) #define maxm (10005) #define mod 1000000007 #define ull unsigned long long #define repne(x,y,i) for(i=(x);i<(y);++i) #define repe(x,y,i) for(i=(x);i<=(y);++i) #define repde(x,y,i) for(i=(x);i>=(y);--i) #define repdne(x,y,i) for(i=(x);i>(y);--i) #define ri register int inline void Read(int &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if(chr=='-')sign=-1; for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;} inline void Read(ll &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if (chr=='-')sign=-1; for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;} ll g[1005],sx[1005],sy[1005],ex[1005],ey[1005]; set<pair<ll,ll> >se[1005]; ll gcd(ll x,ll y){ return (y==0)?x:gcd(y,x%y); } int main() { freopen("a.in","r",stdin); freopen("b.out","w",stdout); //对于每一条线段的每一个整数点可由二元组(sx+k*(ex-sx)/gcd(ex-sx,sy-ey),sy+k*(ey-sy)/gcd(ex-sx,sy-ey))得到 //由此可得到线段中所有的点个数为sum((ex-sx)/gcd(ex-sx,sy-ey)+1) //由于存在重复点需要减去重复点的重复个数 //枚举解方程,若有解则可以确定此点的位置,由于n条线段最多产生n*(n-1)/2个交点 int n; ri i,j,k; Read(n); repe(1,n,i) Read(sx[i]),Read(sy[i]),Read(ex[i]),Read(ey[i]),g[i] = gcd(abs(ex[i]-sx[i]),abs(sy[i]-ey[i])); ll ans = 0; repe(1,n,i) ans = ans + g[i] + 1; //sx[i] + s*(ex[i]-sx[i])/g[i] = sx[j] + t*(ex[j]-sx[j])/g[j] //sy[i] + s*(ey[i]-sy[i])/g[i] = sy[j] + t*(ey[j]-sy[j])/g[j] repe(1,n,i){ ll a = (ex[i] - sx[i])/g[i],b = (ey[i] - sy[i])/g[i]; ll lm = a / gcd(abs(a),abs(b)) * b; repe(i+1,n,j){ ll c = (ex[j]-sx[j])/g[j],d = (ey[j]-sy[j])/g[j]; if(a == 0){ if(c != 0){ if((sx[i] - sx[j]) % c == 0){ ll t = (sx[i] - sx[j]) / c,x = sx[j] + t * c,y = sy[j] + t * d; if((x - sx[i])*(ex[i]-sx[i]) >= 0 && abs(x-sx[i])<=abs(ex[i]-sx[i]) && (x - sx[j])*(ex[j]-sx[j]) >= 0 && abs(x-sx[j])<=abs(ex[j]-sx[j]) && (y - sy[i])*(ey[i]-sy[i]) >= 0 && abs(y-sy[i])<=abs(ey[i]-sy[i]) && (y - sy[j])*(ey[j]-sy[j]) >= 0 && abs(y-sy[j])<=abs(ey[j]-sy[j])){ se[i].insert(make_pair(x,y)); // --ans; } } } continue; } if(b == 0){ if(d != 0){ if((sy[i] - sy[j]) % d == 0){ ll t = (sy[i] - sy[j]) / d,x = sx[j] + t * c,y = sy[j] + t * d; if((x - sx[i])*(ex[i]-sx[i]) >= 0 && abs(x-sx[i])<=abs(ex[i]-sx[i]) && (x - sx[j])*(ex[j]-sx[j]) >= 0 && abs(x-sx[j])<=abs(ex[j]-sx[j]) && (y - sy[i])*(ey[i]-sy[i]) >= 0 && abs(y-sy[i])<=abs(ey[i]-sy[i]) && (y - sy[j])*(ey[j]-sy[j]) >= 0 && abs(y-sy[j])<=abs(ey[j]-sy[j])){ se[i].insert(make_pair(x,y)); // --ans; } } } continue; } if(c * lm / a == d * lm / b) continue; ll tc = c,td = d; c *= lm / a,d *= lm/b; if(c - d == 0) continue; if(((sx[i]-sx[j])*lm/a - (sy[i]-sy[j])*lm/b) % (c-d) != 0) continue; else{ ll t = ((sx[i]-sx[j])*lm/a - (sy[i]-sy[j])*lm/b) / (c-d),c = tc,d = td,x = sx[j] + t * c,y = sy[j] + t * d; if((x - sx[i])*(ex[i]-sx[i]) >= 0 && abs(x-sx[i])<=abs(ex[i]-sx[i]) && (x - sx[j])*(ex[j]-sx[j]) >= 0 && abs(x-sx[j])<=abs(ex[j]-sx[j]) && (y - sy[i])*(ey[i]-sy[i]) >= 0 && abs(y-sy[i])<=abs(ey[i]-sy[i]) && (y - sy[j])*(ey[j]-sy[j]) >= 0 && abs(y-sy[j])<=abs(ey[j]-sy[j])){ se[i].insert(make_pair(x,y)); //--ans; } } } } for(int i = 1;i <= n;++i) ans -= (int)se[i].size(); cout << ans << endl; return 0; }