【poj3714】 Raid

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http://poj.org/problem?id=3714 (题目链接)

现在才搞平面最近点对。。感觉有点尴尬

题意：给出平面上两组点，每组n个，求两组点之间最短距离

Solution1
　　平面最近点对，分治即可。
　　将点按横坐标排序，然后每次二分成左边和右边分别计算最小距离，再计算中间的最小距离，这里需要把中间符合条件的点按照纵坐标排序，然后当当前枚举的两点的纵坐标之差大于答案时break，否则会TLE。
　　
代码：

// poj3714#include<algorithm>#include<iostream>#include<cstring>#include<cstdlib>#include<cstdio>#include<cmath>#include<vector>#define inf 2147483640#define LL long long#define free(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout);using namespace std;inline LL getint() {    LL x=0,f=1;char ch=getchar();    while (ch>'9' || ch<'0') {if (ch=='-') f=-1;ch=getchar();}    while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}    return x*f;}const int maxn=1000010;struct point {double x,y;int flag;}p[maxn];int n,tmp[maxn];bool cmpx(point a,point b) {    return a.x==b.x ? a.y<b.y : a.x<b.x;}bool cmpy(int a,int b) {    return p[a].y==p[b].y ? p[a].x<p[b].x : p[a].y<p[b].y;}double dis(point a,point b) {    return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double solve(int l,int r) {    double res=1e60;    if (l==r) return res;    if (l+1==r) {        if (p[l].flag==p[r].flag) return res;        return dis(p[l],p[r]);    }    int mid=(l+r)>>1;    res=solve(l,mid);    res=min(res,solve(mid+1,r));    int num=0;    for (int i=l;i<=r;i++)        if (fabs(p[i].x-p[mid].x)<=res) tmp[++num]=i;    sort(tmp+1,tmp+num+1,cmpy);    for (int i=1;i<=num;i++)        for (int j=i+1;j<=num;j++) {            if (fabs(p[tmp[i]].y-p[tmp[j]].y)>=res) break; //剪枝            if (p[tmp[i]].flag!=p[tmp[j]].flag) res=min(res,dis(p[tmp[i]],p[tmp[j]]));        }    return res;}int main() {    int T;    scanf("%d",&T);    while (T--) {        scanf("%d",&n);        for (int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y),p[i].flag=0;        for (int i=1;i<=n;i++) scanf("%lf%lf",&p[i+n].x,&p[i+n].y),p[i+n].flag=1;        n<<=1;        sort(p+1,p+1+n,cmpx);        printf("%.3f\n",solve(1,n));    }    return 0;}

Solution2
　　hzwer上惊现平面最近点对的随机化算法（貌似是随机分块），于是我就蒯了过来，虽然并不知道为什么可以这样写，但是好像很厉害的样子。
　　上网搜了下，发现期望复杂度是O（n）的。度娘链接
　　然而= =：
　　这里写图片描述
　　
　　比分治还跑的慢，坑比东西。

代码：

// poj3714#include<algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<cmath>#define LL long long#define inf 2147483640#define Pi acos(-1.0)#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);using namespace std;inline LL getint() {    int f,x=0;char ch=getchar();    while (ch<='0' || ch>'9') {if (ch=='-') f=-1;else f=1;ch=getchar();}    while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}    return x*f;}const int maxn=1000010;struct point {double x,y;int flag;}p[maxn];int n,block,m;bool cmp(point a,point b) {    return a.x==b.x ? a.y<b.y : a.x<b.x;}point rotate(point a,double x) {    return (point){(double)a.x*cos(x)-(double)a.y*sin(x),(double)a.y*cos(x)+(double)a.x*sin(x),a.flag};}double dis(point a,point b) {    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int main() {    int T;scanf("%d",&T);    while (T--) {        scanf("%d",&n);        for (int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y),p[i].flag=0;        for (int i=1;i<=n;i++) scanf("%lf%lf",&p[i+n].x,&p[i+n].y),p[i+n].flag=1;        n<<=1;        block=(int)sqrt(n);        m=n/block+(n%block!=0);        double t=rand()/10000;        for (int i=1;i<=n;i++) p[i]=rotate(p[i],t);        sort(p+1,p+1+n,cmp);        double ans=1e60;        for (int i=1;i<=m;i++) {            int t1=block*(i-1),t2=min(block*i,n);            for (int j=t1;j<=t2;j++)                for (int k=t1+1;k<=t2;k++) if (p[j].flag!=p[k].flag) ans=min(ans,dis(p[j],p[k]));        }        printf("%.3f\n",ans);    }    return 0;}

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