Solution to a problem P2981 [[USACO10FEB] Cows ice Cows on Ice]

Upstairs are looking for ideas to jump from one rock to another rock path, but in fact, this problem can do a virtual node, respectively, for four directions, then simply left to find a virtual node virtual nodes on the line

The question is: how do not use set ???

In fact, you can find a vector-half can achieve the same operation (Although not much longer but also a variety of wa)

The method of FIG stored vector:

They were able to save the x and y directions of the stone, and then stored with vec_x four directions and a virtual node vec_y

As a lazy, negative number is the greatest torment. Each point will add 1e9 can effectively avoid the negative

Then, once for each direction of the judge left up to see how far here is up to judge

long long find_up(long long x, long long y){
  string going = to_string(x);
  if (!mp_horizontal[going].size()) return -1; //这里面没东西
  if (!sorted_hori[going]) {sort(mp_horizontal[going].begin(),mp_horizontal[going].end());sorted_hori[going] = true;}//先把他排序才能二分
  long long le = 0, ri = mp_horizontal[going].size()-1;
  while(le<ri-1){
    long long mid = (le+ri)/2;
    if (mp_horizontal[going][mid]<=y) le = mid;
    else ri = mid;
  }//裸的二分
  if (mp_horizontal[going][le]>y && !between_y(mp_horizontal[going][le],y,going)) return mp_horizontal[going][le];
  if (mp_horizontal[going][ri]>y && !between_y(mp_horizontal[going][ri],y,going)) return mp_horizontal[going][ri];
  return -1;//between_y是查找两点之间是否有石头
}

Now Laijiangjiang between_y.between_y significance is confirmed there will be no stone between virtual nodes. Here, we use the same dichotomy to achieve
the following is between_y, between_x empathy

bool between_y(long long l, long long r, string ptr){
  string going = ptr;
  if (!stone_x[going].size()) return false;//没东西
  if (l>r) swap(l,r);
  if (!sorted_x[going]) {sort(stone_x[going].begin(),stone_x[going].end());sorted_x[going] = true;}
  long long le = 0, ri = stone_x[going].size()-1;

  while(le<ri-1){
    long long mid = (le+ri)/2;

    if (between(l,r,stone_x[going][mid])) return true;
    if (r>stone_x[going][mid]) le = mid;//二分
    else ri = mid;
  }
  return between(l,r,stone_x[going][le])||between(l,r,stone_x[going][ri]);//between定义为比left大且比right小
}

Note that this problem is due to the map and vector memory map, unordered_map is also a good choice

Complete code:

#include <iostream>
#include <algorithm>
#include <unordered_map>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
#define pp pair<long long,long long>
#define f first
#define s second
long long n,sx,sy,ex,ey;
vector<int> vec_x,vec_y,st_x,st_y;
unordered_map<string, bool> sorted_x,sorted_y,sorted_hori,sorted_vert;
unordered_map<string,vector<long long> > mp_vertical,mp_horizontal,stone_x,stone_y;
unordered_map<string,long long> vis;
bool between(long long x, long long y, long long ptr){
  return x<=ptr && y>=ptr;
}//就是懒得每次都这么打
bool between_x(long long l, long long r, string ptr){
  string going = ptr;
  if (!stone_y[going].size()) return false;
  if (l>r) swap(l,r);
  if (!sorted_y[going]) {sort(stone_y[going].begin(),stone_y[going].end());sorted_y[going] = true;}
  long long le = 0, ri = stone_y[going].size()-1;
  while(le<ri-1){
    long long mid = (le+ri)/2;
    if (between(l,r,stone_y[going][mid])) return true;
    if (r<stone_y[going][mid]) ri = mid;
    else le = mid;
  }
  return between(l,r,stone_y[going][le])||between(l,r,stone_y[going][ri]);
}//注意between_x是指y不变,所以这里要用stone_y.
bool between_y(long long l, long long r, string ptr){
  string going = ptr;
  if (!stone_x[going].size()) return false;
  if (l>r) swap(l,r);
  if (!sorted_x[going]) {sort(stone_x[going].begin(),stone_x[going].end());sorted_x[going] = true;}
  long long le = 0, ri = stone_x[going].size()-1;

  while(le<ri-1){
    long long mid = (le+ri)/2;

    if (between(l,r,stone_x[going][mid])) return true;
    if (r>stone_x[going][mid]) le = mid;
    else ri = mid;
  }
  return between(l,r,stone_x[going][le])||between(l,r,stone_x[going][ri]);
}//y方向查找
long long find_up(long long x, long long y){
  string going = to_string(x);
  if (!mp_horizontal[going].size()) return -1;
  if (!sorted_hori[going]) {sort(mp_horizontal[going].begin(),mp_horizontal[going].end());sorted_hori[going] = true;}
  long long le = 0, ri = mp_horizontal[going].size()-1;
  while(le<ri-1){
    long long mid = (le+ri)/2;
    if (mp_horizontal[going][mid]<=y) le = mid;
    else ri = mid;
  }
  if (mp_horizontal[going][le]>y && !between_y(mp_horizontal[going][le],y,going)) return mp_horizontal[going][le];
  if (mp_horizontal[going][ri]>y && !between_y(mp_horizontal[going][ri],y,going)) return mp_horizontal[going][ri];
  return -1;
}//往上
long long find_down(long long x, long long y){
  string going = to_string(x);
  if (!mp_horizontal[going].size()) return -1;
  if (!sorted_hori[going]) {sort(mp_horizontal[going].begin(),mp_horizontal[going].end());sorted_hori[going] = true;}
  long long le = 0, ri = mp_horizontal[going].size()-1;
  while(le<ri-1){
    long long mid = (le+ri)/2;
    if (mp_horizontal[going][mid]>=y) ri = mid;
    else le = mid;
  }
  if (mp_horizontal[going][ri]< y && !between_y(mp_horizontal[going][ri],y,going)) return mp_horizontal[going][ri];
  if (mp_horizontal[going][le]<y  && !between_y(mp_horizontal[going][le],y,going)) return mp_horizontal[going][le];
  return -1;
}//往下
long long find_left(long long x, long long y){
  string going = to_string(y);
  if (!mp_vertical[going].size()) return -1;
  if (!sorted_vert[going]) {sort(mp_vertical[going].begin(),mp_vertical[going].end());sorted_vert[going] = true;}
  long long le = 0, ri = mp_vertical[going].size()-1;
  while(le<ri-1){
    long long mid = (le+ri)/2;
    if (mp_vertical[going][mid]>=x) ri = mid;
    else le = mid;
  }
  if (mp_vertical[going][ri]<x && !between_x(x,mp_vertical[going][ri],going)) return mp_vertical[going][ri];
  if (mp_vertical[going][le]<x && !between_x(x,mp_vertical[going][le],going)) return mp_vertical[going][le];
  return -1;
}//往左
long long find_right(long long x, long long y){
  string going = to_string(y);
  if (!mp_vertical[going].size()) return -1;
  if (!sorted_vert[going]) {sort(mp_vertical[going].begin(),mp_vertical[going].end());sorted_vert[going] = true;}
  long long le = 0, ri = mp_vertical[going].size()-1;
  while(le<ri-1){
    long long mid = (le+ri)/2;
    if (mp_vertical[going][mid]<=x) le = mid;
    else ri = mid;
  }
  if (mp_vertical[going][le]>x && !between_x(x,mp_vertical[going][le],going)) return mp_vertical[going][le];
  if (mp_vertical[going][ri]>x && !between_x(x,mp_vertical[going][ri],going)) return mp_vertical[going][ri];
  return -1;
}//往右
void find_all(int x ,int y){
  cout << find_up(x,y) << " " << find_down(x,y) << " " << find_left(x,y) << " " << find_right(x,y) << endl;
}//这是我自己用来测试的程序
int main(){
  scanf("%lld%lld%lld%lld%lld",&n,&sx,&sy,&ex,&ey);
  sx+=1e9;sy+=1e9;ex+=1e9;ey+=1e9;
  for (long long i=0;i<n;i++){
    long long a,b;scanf("%lld%lld",&a,&b);
    a+=1e9;b+=1e9;
    stone_x[to_string(a)].push_back(b);
    stone_y[to_string(b)].push_back(a);//将石头的状态分别计入map中
    if (a-1>=0) mp_vertical[to_string(b)].push_back(a-1);
    mp_vertical[to_string(b)].push_back(a+1);vec_x.push_back(a+1);
    if (b-1>=0) mp_horizontal[to_string(a)].push_back(b-1);
    mp_horizontal[to_string(a)].push_back(b+1);
    //建立虚拟节点(其实a-1这些不要都无所谓了)
  }
  queue<pp> q;
  q.push(make_pair(sx,sy));
  vis[to_string(sx)+"?"+to_string(sy)] = 0;
  while(!q.empty()){
    long long qf = q.front().f, qs = q.front().s; q.pop();
    long long prev = vis[to_string(qf)+"?"+to_string(qs)];
    if (qf==ex && qs==ey) {cout << prev;return 0;}
    long long le = find_left(qf,qs), ri = find_right(qf,qs), up = find_up(qf,qs), down = find_down(qf,qs);
    if (le!=-1){
      if (!vis[to_string(le)+"?"+to_string(qs)] && (le!=sx || qs!=sy)){
        vis[to_string(le)+"?"+to_string(qs)] = prev+1;
        q.push(make_pair(le,qs));
      }//能往左并且左边的点没拿过
    }
    if (ri!=-1){
      if (!vis[to_string(ri)+"?"+to_string(qs)] && (ri!=sx || qs!=sy)){
        vis[to_string(ri)+"?"+to_string(qs)] = prev+1;
        q.push(make_pair(ri,qs));
      }//能往右并且右边的点没拿过
    }
    if (up!=-1){
      if (!vis[to_string(qf)+"?"+to_string(up)] && (qf!=sx || up!=sy)){
        vis[to_string(qf)+"?"+to_string(up)] = prev+1;
        q.push(make_pair(qf,up));
      }//能往上并且上边的点没拿过
    }
    if (down!=-1){
      if (!vis[to_string(qf)+"?"+to_string(down)] && (qf!=sx || down!=sy)){
        vis[to_string(qf)+"?"+to_string(down)] = prev+1;
        q.push(make_pair(qf,down));
      }//能往下并且下面的点没拿过
    }
  }
}

In fact, this sense code is not great (after all, really long and boring). This is a reference, if the set will use the recommended method upstairs.

Incidentally, leaving a group of cancer data for the benefit of future generations

/*
9
0 0 -2 -3
-3 0
-2 -2
-3 -3
-1 -3
-1 -4
-2 -4
-3 -4
-2 -5
0 -3*/

ps This set of data is no solution (although the original title will not arise, but you can test it)