POJ 2653 Pick-up sticks 線段相交問題

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include <queue>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8const int maxn = 100000+3;struct Point{    double x;    double y;};struct Line{    Point u;    Point v;}stick[maxn];int visit[1000];int Sig(double x){    return (x>eps) - (x<-eps);}double Mult(Point p0, Point p1, Point p2){    return (p1.x-p0.x)*(p2.y-p0.y) - (p2.x-p0.x)*(p1.y-p0.y);}int line_in_line(Line L1, Line L2)  //線段相交 (包含交點){    if(Sig(Mult(L1.u,L2.u,L1.v))*Sig(Mult(L1.u,L2.v,L1.v)) <= 0 && Sig(Mult(L2.u,L1.u,L2.v))*Sig(Mult(L2.u,L1.v,L2.v)) <= 0)        return 1;    return 0;}int main(){#ifndef ONLINE_JUDGE    freopen("in","r",stdin);#endif    int n,top;    while(scanf("%d",&n) && n)    {        scanf("%lf%lf%lf%lf",&stick[1].u.x,&stick[1].u.y,&stick[1].v.x,&stick[1].v.y);        visit[top = 1] = 1;        for(int i = 2; i <= n; i++)        {            scanf("%lf%lf%lf%lf",&stick[i].u.x,&stick[i].u.y,&stick[i].v.x,&stick[i].v.y);            for(int j = 1; j <= top; j++)            {                if(line_in_line(stick[i],stick[visit[j]]))                {                   visit[j] = -1;                }            }            int cur = 0;            for(int j = 1; j <= top; j++)            {                if(visit[j] > 0)                    visit[++cur] = visit[j];            }            top = cur;            visit[++top] = i;        }        printf("Top sticks:");        for(int i = 1; i< top; i++)        {            printf(" %d,",visit[i]);        }        printf(" %d.\n",visit[top]);    }}

line_in_line()函數也可以這麼寫:

int line_in_line(Line L1, Line L2)  //線段相交 (包含交點){//    矩陣跨立    return (max(L1.u.x, L1.v.x) >= min(L2.u.x, L2.v.x) &&            max(L2.u.x, L2.v.x) >= min(L1.u.x, L1.v.x) &&            max(L1.u.y, L1.v.y) >= min(L2.u.y, L2.v.y) &&            max(L2.u.y, L2.v.y) >= min(L1.u.y, L1.v.y) &&            Mult(L1.u,L2.v,L2.u)*Mult(L2.v,L1.v,L2.u)>=0 &&            Mult(L2.u,L1.v,L1.u)*Mult(L1.v,L2.v,L1.u)>=0) ;}

我奇怪之前我寫的為什麼會逾時.

代碼.我用判斷visit最佳化了.知道的給個回複:

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include <queue>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8const int maxn = 100000+3;struct Point{    double x;    double y;};struct Line{    Point u;    Point v;}stick[maxn];bool visit[maxn];int Sig(double x){    return (x>eps) - (x<-eps);}double Mult(Point p0, Point p1, Point p2){    return (p1.x-p0.x)*(p2.y-p0.y) - (p2.x-p0.x)*(p1.y-p0.y);}int line_in_line(Line L1, Line L2){    if(Sig(Mult(L1.u,L2.u,L1.v))*Sig(Mult(L1.u,L2.v,L1.v))<=0 && Sig(Mult(L2.u,L1.u,L2.v))*Sig(Mult(L2.u,L1.v,L2.v)) <=0)        return 1;    return 0;}int main(){#ifndef ONLINE_JUDGE    freopen("in","r",stdin);#endif    int n;    while(scanf("%d",&n) && n)    {        memset(visit, false, sizeof(visit));        for(int i = 1; i <= n; i++)        {            scanf("%lf%lf%lf%lf",&stick[i].u.x,&stick[i].u.y,&stick[i].v.x,&stick[i].v.y);            for(int j = 1; j < i; j++)            {                if(!visit[j] && line_in_line(stick[i],stick[j]))                    visit[j] = true;            }        }        printf("Top sticks:");        for(int i = 1; i< n; i++)        {            if(!visit[i])                printf(" %d,",i);        }        printf(" %d.\n",n);    }}