Codeforces 724C Ray Tracing 模拟

最新推荐文章于 2021-05-28 18:36:07 发布

原创最新推荐文章于 2021-05-28 18:36:07 发布 · 1.8w 阅读

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这篇博客介绍了Codeforces 724C问题的解决方案，涉及光束从(0,0)出发向(1,1)移动的模拟过程。光束遇到墙角会停止，碰到墙面则按入射角等于出射角反射。题目要求计算光束首次经过传感器所在位置的时间，若未经过则输出-1。博主首先分享了错误的尝试，即按点和按边搜索传感器，然后阐述了正确的思路：计算传感器所在直线的截距，并按边搜索，通过边的截距找到对应传感器。博客最后附上了正确代码。" 134498792,20267071,Python随机颜色绘制七段数码管时间代码,"['Python', '开发语言', '图形绘制']

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题意：给你一张n*m的图，横坐标从0到m，纵坐标从0到n，直线y=n，y=0，x=0，x=m都是墙，现有光束从0,0出发向1，1射去，每进过一个点耗费1秒，当遇到墙角时，光束停止，当射到墙面上时，反射，入射角等于出射角，先有k个传感器在不同的点上，问光束第一次经过这些点的时间，若不经过输出-1。

题解：这题真是卡了我半天，真的是半天，我先按点一个一个搜t在test7，再按边搜然后对于每条边找一遍有没有传感器在上面，又t在了8，当然啦，这样的姿势一看就是不正确的，然而蒟蒻的我还是连踩两发坑，那么正确的思路应该是先求出每个传感器所在直线y=x+b,y=-x+b的截距。然后按边搜，找出每条边的截距，然后再通过截距找对应的传感器.

下面是正确的代码。

#include<bits/stdc++.h>
#define MEM(a,x) memset(a,x,sizeof(a));
#define MEMINF(a) memset(a,0x3f,sizeof(a));
using namespace std;
typedef long long LL;
const int MAXN=1e5+10;
const LL INF=0x3f3f3f3f3f3f3f3fll;
struct node {
  int x,y;
  LL step;
  int dir;
  int id;
};
int vis[MAXN];
typedef pair<int,int> PAIR;
LL ans[MAXN];
map<int,vector<pair<int,int> > >sz,sf;
void judge(node now) {
  int b;
  int flag=0;
  if (now.dir==1||now.dir==2) b=now.y-now.x;
  else b=now.x+now.y,flag=1;
  if (flag==1) {
    for (auto it :sf[b]) {
      int rx=it.first,id=it.second;
      ans[id]=min(ans[id],now.step+abs(now.x-rx));
    }
    sf.erase(b);
  }
  else {
    for (auto it : sz[b]) {
      int rx=it.first,id=it.second;
      ans[id]=min(ans[id],now.step+abs(now.x-rx));
    }
    sz.erase(b);
  }
}


int main() {
  int n,m,k;
  scanf("%d %d %d",&m,&n,&k);
  int x,y;
  for (int i=1; i<=k; ++i) {
    scanf("%d %d",&x,&y);
    swap(x,y);
    sz[y-x].push_back(PAIR(x,i));
    sf[y+x].push_back(PAIR(x,i));
  }
  MEMINF(ans);
  node now;
  now.x=0,now.y=0,now.dir=1,now.step=0;
  int mx,my;
  bool flag=false;
  int cnt=0;
  while (true) {
    node lim;
    if (now.dir==1) {
      int r=m-now.y,u=n-now.x;
      if (u<r) lim.dir=3,mx=u,my=u;
      else if(r<u)lim.dir=0,mx=r,my=r;
      else if (r==u) mx=my=r,flag=true;

    }
    else if(now.dir==0) {
      int l=now.y-0,u=n-now.x;
      if (l<u) mx=l,my=-l,lim.dir=1;
      else if(u<l)mx=u,my=-u,lim.dir=2;
      else mx=l,my=-l,flag=true;
    }
    else if(now.dir==2) {
      int d=now.x-0,l=now.y-0;
      if (l<d) mx=my=-l,lim.dir=3;
      else if(d<l)mx=my=-d,lim.dir=0;
      else mx=my=-d,flag=true;
    }
    else if(now.dir==3) {
      int r=m-now.y,d=now.x-0;
      if (r<d) mx=-r,my=r,lim.dir=2;
      else if(d<r)mx=-d,my=d,lim.dir=1;
      else mx=-d,my=d,flag=true;
    }
        lim.x=now.x+mx;
        lim.y=now.y+my;
        lim.step=now.step+abs(mx);
        judge(now);
        now=lim;
       if(now.x==0&&now.y==0) break;
       else if (now.x==0&&now.y==m) break;
       else if (now.x==n&&now.y==m) break;
       else if (now.x==n&&now.y==0) break;
  }
  for (int i=1; i<=k; ++i) {
    ans[i]==INF?puts("-1"):printf("%lld\n",ans[i]);
  }
}

接下来是两发t的，写了比较长时间不舍得删掉，哈哈。

<span style="color:#000000;">#include<bits/stdc++.h>
#define MEM(a,x) memset(a,x,sizeof(a));
#define MEMINF(a) memset(a,0x3f,sizeof(a));
using namespace std;
typedef long long LL;
const int MAXN=1e5+10;
struct node {
  int x,y;
  int step;
  int dir;
  int id;
};
int dx[4]={1,1,-1,-1};
template<class T>
inline bool read(T &n)
{
    T x = 0, tmp = 1;
    char c = getchar();
    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
    if(c == EOF) return false;
    if(c == '-') c = getchar(), tmp = -1;
    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
    n = x*tmp;
    return true;
}
template <class T>
inline void write(T n)
{
    if(n < 0)
    {
        putchar('-');
        n = -n;
    }
    int len = 0,data[20];
    while(n)
    {
        data[len++] = n%10;
        n /= 10;
    }
    if(!len) data[len++] = 0;
    while(len--) putchar(data[len]+48);
}
int dy[4]={-1,1,-1,1};

int vis[MAXN];
typedef pair<int,int> PAIR;
typedef map<PAIR,int> mpp;
mpp mp;
queue<node>q;
int qsize;
double k1,k2;

void judge(node u,node lim) {
  int k=qsize;
  for(int i=0; i<k; ++i){ 
    node v=q.front();
    q.pop();
    k2=((double)(lim.x-u.x))/((double)(lim.y-u.y));
    double left=double(v.y-u.y);
    double right=double(v.x-u.x)*k2;
    double aa=left-right;
    
    if (aa==0&&!vis[v.id]) {
      vis[v.id]=abs(v.x-u.x)+u.step,qsize--;
    }
    else q.push(v);
  }
}



int main() {
  int n,m,k;
  read(m),read(n),read(k);
  int x,y;
  for (int i=1; i<=k; ++i) {
    read(x),read(y);
    node v;
    v.x=y,v.y=x;
    v.id=i;
    q.push(v);
    qsize++;
  }
  node now;
  now.x=1,now.y=1,now.dir=1,now.step=1;
  int mx,my;
  bool flag=false;
  int cnt=0;
  while (true) {
    if (!qsize) break;
    cnt++;
    node lim;
    if (cnt>1e5) break;
    if (now.dir==1) {
      int r=m-now.y,u=n-now.x;
      if (u<r) lim.dir=3,mx=u,my=u;
      else if(r<u)lim.dir=0,mx=r,my=r;
      else if (r==u) mx=my=r,flag=true;

    }
    else if(now.dir==0) {
      int l=now.y-0,u=n-now.x;
      if (l<u) mx=l,my=-l,lim.dir=1;
      else if(u<l)mx=u,my=-u,lim.dir=2;
      else mx=l,my=-l,flag=true;
    }
    else if(now.dir==2) {
      int d=now.x-0,l=now.y-0;
      if (l<d) mx=my=-l,lim.dir=3;
      else if(d<l)mx=my=-d,lim.dir=0;
      else mx=my=-d,flag=true;
    }
    else if(now.dir==3) {
      int r=m-now.y,d=now.x-0;
      if (r<d) mx=-r,my=r,lim.dir=2;
      else if(d<r)mx=-d,my=d,lim.dir=1;
      else mx=-d,my=d,flag=true;
    }
        lim.x=now.x+mx;
        lim.y=now.y+my;
        lim.step=now.step+abs(mx);
        judge(now,lim);
        now=lim;
       if(now.x==0&&now.y==0) break;
       else if (now.x==0&&now.y==m) break;
       else if (now.x==n&&now.y==m) break;
       else if (now.x==n&&now.y==0) break;
  }
  for (int i=1; i<=k; ++i) {
    if (vis[i]) write(vis[i]),puts("");
    else puts("-1");
  }
}

</span>

<span style="color:#000000;">#include<bits/stdc++.h>
#define MEM(a,x) memset(a,x,sizeof(a));
#define MEMINF(a) memset(a,0x3f,sizeof(a));
using namespace std;
typedef long long LL;
const int MAXN=1e5+10;
struct node {
  int x,y;
  int step;
  int dir;
};
int dx[4]={1,1,-1,-1};
int dy[4]={-1,1,-1,1};

int ans[MAXN];
bool vis[MAXN];
typedef pair<int,int> PAIR;
typedef map<PAIR,int> mpp;
mpp mp;

int main() {
  int n,m,k;
  scanf("%d %d %d",&n,&m,&k);
  int x,y;
  for (int i=1; i<=k; ++i) {
    scanf("%d %d",&x,&y);
    mp[PAIR(x,y)]=i;
  }
  node pre,now;
  MEM(vis,0);
  MEM(ans,0);
  now.x=1,now.y=1,now.step=1,now.dir=1;
  int flag=0;

  int cnt=0;
  
  while(true) {
    mpp::iterator it; 
    it=mp.find(PAIR(now.x,now.y));
    int v=it->second;
    if (v&&!vis[v]) {
      ans[v]=now.step;
      vis[v]=1;
      flag++;
    }
    if (cnt>1e5)break;
    if (flag==k) break;
    now.x+=dx[now.dir];
    now.y+=dy[now.dir];
    now.step++;
    if (now.x==0&&now.y==0)break;
    if (now.x==n&&now.y==m) break;
    if (now.x==0&&now.y==m) break;
    if (now.x==n&&now.y==m)break;

    if (now.x==0) {
      if (now.dir==3) now.dir=1;
      else if (now.dir==2) now.dir=0;
    }
    else if (now.x==n) {
      if (now.dir==1) now.dir=3;
      else if (now.dir==0) now.dir=2;
    }
    else if (now.y==0) {
      if (now.dir==2) now.dir=3;
      else if (now.dir==0) now.dir=1;
    }
    else if (now.y==m) {
      if (now.dir==3) now.dir=2;
      else if (now.dir==1) now.dir=0;
    }
  }
  for (int i=1; i<=k; ++i) {
    if (ans[i]==0)puts("-1");
    else printf("%d\n",ans[i]);
  }
}

</span>