【AHOI2005】穿越磁场

题面

https://www.luogu.org/problem/P2537

题解

#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
#define ri register int
#define N 105
using namespace std;
const int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int n,c;
int sx,sy,tx,ty,cx,cy;
int x1[N],y1[N],x2[N],y2[N];
int nx1[N],ny1[N],nx2[N],ny2[N];
int tr1[N<<1],tr2[N<<1];

struct node {
  int x; bool s;
  int id;
  bool operator < (const node &rhs) const {
    return x<rhs.x;
  }
} a[N<<1];

bool in[N][2*N][2*N];
bool vis[2*N][2*N];

struct nod {
  int x,y,s;
};
deque<nod> q;

void get(int &a,int &b,int x,int y){
  for (ri i=1;i<cx;i++) 
    for (ri j=1;j<cy;j++) if (tr1[i]<=x && x<=tr1[i+1] && tr2[j]<=y && y<=tr2[j+1]) {
      a=i;
      b=j;
      return;
    }
}

int daijia(int x1,int y1,int x2,int y2){
  for (ri i=1;i<=n;i++) {
    if ((in[i][x1][y1] && !in[i][x2][y2])||(!in[i][x1][y1] && in[i][x2][y2])) return 1;
  }
  return 0;
}

int main(){
  scanf("%d",&n);
  for (ri i=1;i<=n;i++) {
    scanf("%d %d %d",&x1[i],&y1[i],&c);
    x2[i]=x1[i]+c; y2[i]=y1[i]+c;
  }
  scanf("%d %d %d %d",&sx,&sy,&tx,&ty);
  for (ri i=1;i<=n;i++) a[2*i-1]=(node){x1[i],0,i},a[2*i]=(node){x2[i],1,i};
  n++; a[2*n-1]=(node){-1,0,n}; a[2*n]=(node){100000,0,n}; 
  sort(a+1,a+2*n+1);
  cx=0;
  for (ri i=1;i<=2*n;i++) {
    if (a[i].x!=a[i-1].x) {
      cx++;
      tr1[cx]=a[i].x;
    }
    if (!a[i].s) nx1[a[i].id]=cx; else nx2[a[i].id]=cx;
  }
  //for (ri i=1;i<=cx;i++) cout<<tr1[i]<<" "; cout<<endl;
  n--;
  for (ri i=1;i<=n;i++) a[2*i-1]=(node){y1[i],0,i},a[2*i]=(node){y2[i],1,i};
  n++; a[2*n-1]=(node){-1,0,n}; a[2*n]=(node){100000,0,n};
  sort(a+1,a+2*n+1);
  cy=0;
  for (ri i=1;i<=2*n;i++) {
    if (a[i].x!=a[i-1].x) {
      cy++;
      tr2[cy]=a[i].x;
    }
    if (!a[i].s) ny1[a[i].id]=cy; else ny2[a[i].id]=cy;
  }
  n--;

  //for (ri i=1;i<=cy;i++) cout<<tr2[i]<<" "; cout<<endl;

  //puts("83");
  //for (ri i=1;i<=n;i++) printf("%d %d %d %d
",nx1[i],nx2[i],ny1[i],ny2[i]);
  for (ri i=1;i<=n;i++)
    for (ri x=1;x<cx;x++)
      for (ri y=1;y<cy;y++) if (nx1[i]<=x && x+1<=nx2[i] && ny1[i]<=y && y+1<=ny2[i]) {
        in[i][x][y]++;
        //cout<<"87 "<<i<<" "<<x<<" "<<y<<endl;
      }
  
  //puts("80");
  int fx,fy,ex,ey;
  get(fx,fy,sx,sy);
  get(ex,ey,tx,ty);
  //cout<<fx<<" "<<fy<<endl;
  //cout<<ex<<" "<<ey<<endl;
  q.push_back((nod){fx,fy,0}); vis[fx][fy]=1;
  while (!q.empty()) {
    nod cur=q.front(); q.pop_front();
    for (ri i=0;i<4;i++) {
      int nx=cur.x+dx[i],ny=cur.y+dy[i];
      if (nx<cx && ny<cy && nx>=1 && ny>=1 && !vis[nx][ny]) {
        vis[nx][ny]=1;
        int t=daijia(cur.x,cur.y,nx,ny);
        if (!t) q.push_front((nod){nx,ny,cur.s+daijia(cur.x,cur.y,nx,ny)});
        else q.push_back((nod){nx,ny,cur.s+daijia(cur.x,cur.y,nx,ny)});
        if (nx==ex && ny==ey) {
          printf("%d
",cur.s+daijia(cur.x,cur.y,nx,ny));
          return 0;
        }
      }
    }
  }
  puts("-1");
  return 0;
}
原文地址:https://www.cnblogs.com/shxnb666/p/11278457.html