XJOI NOI2015训练题7 题解

最新推荐文章于 2024-09-11 11:00:10 发布

Zeyu_King

最新推荐文章于 2024-09-11 11:00:10 发布

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分类专栏：几何日常 dp&递推文章标签：计算几何递推

本文链接：https://blog.youkuaiyun.com/Zeyu_King/article/details/46485299

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该博客介绍了XJOI NOI2015训练赛中的一道计算几何题目，强调了题目不进行模1e9+7的操作导致需要处理高精度。博主分享了状态转移方程和优化技巧，指出在不超时的前提下，使用long long代替更高精度的数据类型能够解决问题。同时，博主提到了原题来源和解决思路，包括利用凸包和旋转卡壳的方法，并提及预处理关键点的上下凸壳来合并求解。文章还提及了寻找矩形内最高、最低、最左、最右点的二分查找优化方法以及避免错误的数据结构选择导致的性能问题。

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T1、

~~原本以为是水题的，但是题目里有这么一句话”不模1e9+7“。。。。~~

可以注意到上下左右都要求单调增，那么状态就很好确定了

f[i][j]表示第一行放了前i个格子,第二行放了前j个格子，并且这些数都是小于等于i+j的

转义很好写：f[i][j]可以推给f[i+1][j]和f[i][j+1]。需要时时保证i>j

另外注意此处格子是否可放。

由于不取模，需要用高精。但是实现只有0.2s

——卡常!

最后还是被卡了20分——本机测试不超时的

订正的时候发现用long long压18位居然就能过了，而9位int死活不过。。。。

还有有些人用这个不算开挂吗？　__attribute__((optimize("O2")))

#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
const long long Mod = (1e18);
int a[2][1005];
int n,i,j;
struct BITINT{
  int h;
  long long a[40];
  BITINT operator +(const BITINT &x) const
  {
    if (h==0) return x;
    BITINT ret;
    memset(ret.a,0,sizeof(ret.a));
    ret.h = max(h, x.h);
    for (int i=0;i<ret.h||ret.a[i+1]>0;i++)
    {
      ret.a[i] += a[i]+x.a[i];
      while (ret.a[i]>=Mod)
        ret.a[i+1]++, ret.a[i]-=Mod;
    }
    while (ret.a[ret.h]>0) ret.h++;
    //for (int i=ret.h;i<70;i++) ret.a[i]=0;
    return ret;
  }
  void print(){
    if (h==0) {printf("0\n");return;}
    printf("%lld",a[h-1]);
    for (int i=h-2;i>=0;i--){
      long long x = Mod/10;
      if (a[i]==0) {printf("000000000000000000");continue;}
      while (a[i]<x) {printf("0");x/=10;}
      printf("%lld",a[i]);
    }
    printf("\n");
  }
} f[1005][1005];
 
int main(){
  scanf("%d",&n);
  for (i=1;i<=n;i++)
    scanf("%d",&a[0][i]);
  for (i=1;i<=n;i++)
    scanf("%d",&a[1][i]);
  f[0][0].h = f[0][0].a[0] = 1;
 
  for (i=0;i<=n;i++)
    for (j=0;j<=i;j++)
    if (f[i][j].h>0){
      if (i<n && (a[0][i+1]==0 || a[0][i+1]==i+j+1))
        f[i+1][j] = f[i+1][j] + f[i][j];
      if (j<i && (a[1][j+1]==0 || a[1][j+1]==i+j+1))
        f[i][j+1] = f[i][j+1] + f[i][j];
    }
 
  f[n][n].print();
  return 0;
}

T2、

shtsc2011的原题。

凸包＋旋转卡壳，脑补脑补就知道了。。。。

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
typedef long long LL;
const int Maxn = 200005;
int n,m,i,j,upcnt,dwcnt;
int opp[Maxn];
double ans, dis1[Maxn], dis2[Maxn];
struct Point
{
  int x, y;
  void read()
  {
    scanf("%d%d",&x,&y);
  }
  bool operator <(const Point &a)const
  {
    return x<a.x || (x==a.x && y<a.y);
  }
  Point operator *(const int &a)const
  {
    return (Point){x*a, y*a};
  }
  Point operator +(const Point &a)const
  {
    return (Point){x+a.x, y+a.y};
  }
  Point operator -(const Point &a)const
  {
    return (Point){x-a.x, y-a.y};
  }
  double len()
  {
    return sqrt(1.*x*x+1.*y*y);
  }
} P, dot[Maxn], up[Maxn], dw[Maxn];
 
int mult(Point P1,Point P2)
{
  if ((LL)P1.x*P2.y>(LL)P1.y*P2.x) return 1;
  if ((LL)P1.x*P2.y<(LL)P1.y*P2.x) return -1;
  return 0;
}
int mult(Point P0,Point P1,Point P2)
{
  return mult(P1-P0,P2-P0);
}
 
void convex_hull(){
  sort(dot+1,dot+n+1);
  up[1] = dot[1]; up[2] = dot[2]; upcnt = 2;
  for (i=3;i<=n;i++){
    while (upcnt>1 && mult(up[upcnt-1],up[upcnt],dot[i])>=0 ) upcnt--;
    up[++upcnt] = dot[i];
  }
  dw[1] = dot[1]; dw[2] = dot[2]; dwcnt = 2;
  for (i=3;i<=n;i++){
    while (dwcnt>1 && mult(dw[dwcnt-1],dw[dwcnt],dot[i])<=0 ) dwcnt--;
    dw[++dwcnt] = dot[i];
  }
 
  for (i=1;i<upcnt;i++) dot[++m] = up[i];
  for (i=dwcnt;i>1;i--) dot[++m] = dw[i];
}
 
void find_opp(int i,int &j){
  for (;true;j=j%m+1){
    int t1 = mult(dot[i%m+1]-dot[i], dot[j]-dot[j-1]);
    int t2 = mult(dot[i%m+1]-dot[i] ,dot[j%m+1]-dot[j]);
    if (t1*t2<=0) return;
  }
}
 
double MULT(Point P1,Point P2)
{
  return 1.*P1.x*P2.y-1.*P1.y*P2.x;
}
 
double Distance(Point P1,Point P2){
  double ret = MULT(P1,P2);
  if (ret<0) ret = -ret;
  ret /= P1.len();
  return ret;
}
 
void calc(int i,int j1,int j2){
  for (int j=j1;j!=j2%m+1;j=j%m+1){
    Point Q = P*2-(dot[i]+dot[j]);
    int t1 = mult(Q,dot[j]-dot[j-1]) * mult(Q,dot[j%m+1]-dot[j]);
    int t2 = mult(Q,dot[i]-dot[i-1]) * mult(Q,dot[i%m+1]-dot[i]);
    if (t1<=0 && t2<=0){
      double d1 = Distance( Q, dot[i]-P );
      double d2 = Distance( Q, dot[j%m+1]-P );
      double d3 = Distance( Q, dot[(j+1)%m+1]-P );
      ans = fmin( ans, fmax(d1, fmax(d2, d3) ) );
    }
  }
}
 
#define fmin(x,y) ( (x)<(y)?(x):(y) )
#define fmax(x,y) ( (x)>(y)?(x):(y) )
const double Eps = 1e-12;
void work(){
  //for (i=1;i<=m;i++) printf("%d %d\n",dot[i].x,dot[i].y);
  dot[0] = dot[m]; ans = 1e18;
  for (i=1,j=3;i<=m;i++){
    find_opp(i,j); opp[i] = j;
    double t1 = Distance( (dot[i%m+1]-dot[i]), (dot[i]-P) );
    double t2 = Distance( (dot[i%m+1]-dot[i]), (dot[i%m+1]-P) );
    double t3 = Distance( (dot[i%m+1]-dot[i]), (dot[j]-P) );
    double t4 = Distance( (dot[i%m+1]-dot[i]), (dot[j%m+1]-P) );
    if (t2>t1) t1 = t2; if (t4>t3) t3 = t4;
    dis1[i] = t1; dis2[i] = t3;
    ans = fmin(ans, fmax(t1, t3) );
  }
  for (i=1;i<=m;i++){
    if (dis2[i]-dis1[i]<=Eps && dis1[i%m+1]-dis2[i%m+1]<=Eps ) calc(i%m+1,opp[i],opp[i%m+1]);
    if (dis1[i]-dis2[i]<=Eps && dis2[i%m+1]-dis1[i%m+1]<=Eps ) calc(i%m+1,opp[i],opp[i%m+1]);
  }
  ans = (double)floor(ans*1000) / 1000.;
  printf("%.3lf\n",ans);
}
 
int main(){
  //freopen("t2.in","r",stdin);
  //freopen("t2.out","w",stdout);
  scanf("%d",&n);
  P.read();
  for (i=1;i<=n;i++)
    dot[i].read();
  convex_hull();
  work();
  return 0;
}

T3、

压轴神题

思路很好

只要我们知道长方形内最高点，最低点，最左点和最右点，预处理这些点之间的上下凸壳就可以合并求答案了。

找最点可以用二分查找矩形内是否有点在logn的时间内搞定。~~（我会说我之前求矩形内是否有点写的是二维树状数组，平白无故多了两个log然后T得飞起吗？）~~

用f1[i][j]表示在点i和点j围成的矩形内的上凸壳的面积，f2[i][j]表示在点i和点j围成的矩形内的下凸壳的面积，预处理可以做到O(n^2)

代码很呵呵

#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
const int Maxn = 3005;
typedef long long LL;
int n,m,Q,k,i,j,tx,ty;
int a,b,c,d,A,B,C,D;
int nx,ny,upcnt,dwcnt;
int dw[Maxn], up[Maxn];
int ax[Maxn], ay[Maxn];
int sum[Maxn][Maxn];
LL ans, f1[Maxn][Maxn], f2[Maxn][Maxn];
vector <int> ex[Maxn], ey[Maxn];
struct Point
{
  int x, y;
  void read()
  {
    scanf("%d%d",&x,&y);
  }
  bool operator <(const Point &a)const
  {
    return x<a.x || (x==a.x && y<a.y);
  }
  Point operator -(const Point &a)const
  {
    return (Point){x-a.x, y-a.y};
  }
} dot[Maxn], Dot[Maxn];
 
LL mult(Point P0,Point P1)
{
  return (LL)P0.x*P1.y - (LL)P1.x*P0.y;
}
LL mult(Point P0,Point P1,Point P2)
{
  return mult(P1-P0,P2-P0);
}
 
int query(int x1,int x2,int y1,int y2)
{
  return sum[x2][y2] - sum[x1-1][y2] - sum[x2][y1-1] + sum[x1-1][y1-1];
}
 
void init(){
  for (i=1;i<=n;i++){
    up[1] = i; upcnt = 1;
    for (j=i-1;j>0;j--)
    if (dot[i].y>=dot[j].y)
      {up[++upcnt] = j; break;}
    for (j--;j>0;j--)
    if (dot[i].y>=dot[j].y){
      while (upcnt>1 && mult(dot[up[upcnt-1]], dot[up[upcnt]], dot[j])<=0 ) upcnt--;
      f1[j][i] = f1[up[upcnt]][i] + mult(dot[i], dot[up[upcnt]], dot[j]);
      up[++upcnt] = j;
    }
 
    up[1] = i; upcnt = 1;
    for (j=i+1;j<=n;j++)
    if (dot[i].y>=dot[j].y)
      {up[++upcnt] = j; break;}
    for (j++;j<=n;j++)
    if (dot[i].y>=dot[j].y){
      while (upcnt>1 && mult(dot[up[upcnt-1]], dot[up[upcnt]], dot[j])>=0 ) upcnt--;
      f1[i][j] = f1[i][up[upcnt]] - mult(dot[i], dot[up[upcnt]], dot[j]);
      up[++upcnt] = j;
    }
 
    dw[1] = i; dwcnt = 1;
    for (j=i-1;j>0;j--)
    if (dot[i].y<=dot[j].y)
      {dw[++dwcnt] = j; break;}
    for (j--;j>0;j--)
    if (dot[i].y<=dot[j].y){
      while (dwcnt>1 && mult(dot[dw[dwcnt-1]], dot[dw[dwcnt]], dot[j])>=0 ) dwcnt--;
      f2[j][i] = f2[dw[dwcnt]][i] - mult(dot[i], dot[dw[dwcnt]], dot[j]);
      dw[++dwcnt] = j;
    }
 
    dw[1] = i; dwcnt = 1;
    for (j=i+1;j<=n;j++)
    if (dot[i].y<=dot[j].y)
      {dw[++dwcnt] = j; break;}
    for (j++;j<=n;j++)
    if (dot[i].y<=dot[j].y){
      while (dwcnt>1 && mult(dot[dw[dwcnt-1]], dot[dw[dwcnt]], dot[j])<=0 ) dwcnt--;
      f2[i][j] = f2[i][dw[dwcnt]] + mult(dot[i], dot[dw[dwcnt]], dot[j]);
      dw[++dwcnt] = j;
    }
 
  }
}
 
int find_left(int a,int b,int c,int d){
  int L = a, R = b, t=0;
  while (L<=R){
    int mid = (L+R)>>1;
    if (query(a,mid,c,d)==0) L=mid+1;
      else R=mid-1, t=mid;
  }
  L = 0; R = ex[t].size()-1;
  int ret = 0;
  while (L<=R){
    int mid = (L+R)>>1;
    if (dot[ex[t][mid]].y<ay[c]) L=mid+1;
      else ret=ex[t][mid], R=mid-1;
  }
  return ret;
}
 
int find_right(int a,int b,int c,int d){
  int L = a, R = b, t=0;
  while (L<=R){
    int mid = (L+R)>>1;
    if (query(mid,b,c,d)==0) R=mid-1;
      else L=mid+1, t=mid;
  }
  L = 0; R = ex[t].size()-1;
  int ret = 0;
  while (L<=R){
    int mid = (L+R)>>1;
    if (dot[ex[t][mid]].y>ay[d]) R=mid-1;
      else ret=ex[t][mid], L=mid+1;
  }
  return ret;
}
 
int find_down(int a,int b,int c,int d){
  int L = c, R = d, t=0;
  while (L<=R){
    int mid = (L+R)>>1;
    if (query(a,b,c,mid)==0) L=mid+1;
      else R=mid-1, t=mid;
  }
  L = 0; R = ey[t].size()-1;
  int ret=0;
  while (L<=R){
    int mid = (L+R)>>1;
    if (dot[ey[t][mid]].x<ax[a]) L=mid+1;
      else ret=ey[t][mid], R=mid-1;
  }
  return ret;
}
 
int find_up(int a,int b,int c,int d){
  int L = c, R = d, t=0;
  while (L<=R){
    int mid = (L+R)>>1;
    if (query(a,b,mid,d)==0) R=mid-1;
      else L=mid+1, t=mid;
  }
  L = 0; R = ey[t].size()-1;
  int ret=0;
  while (L<=R){
    int mid = (L+R)>>1;
    if (dot[ey[t][mid]].x>ax[b]) R=mid-1;
      else ret=ey[t][mid], L=mid+1;
  }
  return ret;
}
 
int main(){
  scanf("%d%d",&k,&n);
  for (i=1;i<=n;i++){
    dot[i].read();
    ax[i] = dot[i].x;
    ay[i] = dot[i].y;
  }
  sort(dot+1,dot+n+1);
  sort(ax+1,ax+n+1);
  nx = unique(ax+1,ax+n+1)-ax-1;
  sort(ay+1,ay+n+1);
  ny = unique(ay+1,ay+n+1)-ay-1;
 
  for (i=1;i<=n;i++){
    tx = lower_bound(ax+1,ax+nx+1,dot[i].x)-ax;
    ty = lower_bound(ay+1,ay+ny+1,dot[i].y)-ay;
    ex[tx].push_back(i); ey[ty].push_back(i);
    sum[tx][ty]++;
  }
  for (i=1;i<=nx;i++)
    for (j=1;j<=ny;j++)
      sum[i][j] += sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1];
  init();
 
  scanf("%d",&Q);
  while (Q--){
    scanf("%d%d%d%d",&a,&b,&c,&d);
    a = lower_bound(ax+1,ax+nx+1,a)-ax;
    b = upper_bound(ax+1,ax+nx+1,b)-ax-1;
    c = lower_bound(ay+1,ay+ny+1,c)-ay;
    d = upper_bound(ay+1,ay+ny+1,d)-ay-1;
    //printf("%d %d %d %d\n",a,b,c,d);
    if (query(a,b,c,d)<3) {printf("0\n");continue;}
     
    if (a>=b || c>=d) {printf("0\n");continue;}
    A = find_left(a,b,c,d);
    B = find_right(a,b,c,d);
    C = find_down(a,b,c,d);
    D = find_up(a,b,c,d);
    //printf("%d %d %d %d\n",A,B,C,D);
    if (A==B || C==D) {printf("0\n");continue;}
 
    ans = f1[A][D] + f1[D][B] + f2[A][C] +f2[C][B];
    ans += mult(dot[A], dot[C], dot[D]) + mult(dot[D], dot[C], dot[B]);
    printf("%.1lf\n",(double)ans/2.);
  }
  return 0;
}