Bzoj 1967 Ahoi2005 Cross through magnetic field FLOODFILL+BFS

Question: N squares on a given plane, to find at least the number of boundaries through which one point goes to another

At first, I want to determine for each square whether the start and end points are on the same side = = But the inverse is obviously

Consider the n<=100, which can be discretized and then labeled with FloodFill for each block of area

Then run the shortest way. Because the edge is 1, so you can use BFS to find the shortest possible.

Even the edge of the company hung up half a night.

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define M 310using namespace Std;struct abcd{int x,y,c;friend istream& operator >> (IStream &_,abcd &a) {scanf ("%d%d%d", & AMP;A.X,&AMP;A.Y,&AMP;A.C); return _;}} A[m];struct rectangle{int X1,x2,y1,y2;} Rectangles[m];struct edge{int To,next;} Table[500500];int head[m*m],tot;int m=1,n=1,cnt,t;int xa,ya,xb,yb,map[m][m],f[100100];p air<int,int*> B[M]; BOOL map1[m][m],map2[m][m];void Add (int x,int y) {table[++tot].to=y;table[tot].next=head[x];head[x]=tot;} void FloodFill (int x,int y) {int xx,yy;map[x][y]=t;xx=x,yy=y-1;if (!map1[x][y]) {if (y!=1 &&!map[xx][yy]) FloodFill (XX,YY);} else if (Map[xx][yy]) Add (T,map[xx][yy]), add (map[xx][yy],t), Xx=x,yy=y+1;if (!map1[x][y+1]) {if (Y!=n &&!map[ XX][YY]) FloodFill (XX,YY);} else if (Map[xx][yy]) Add (T,map[xx][yy]), add (map[xx][yy],t), Xx=x-1,yy=y;if (!map2[x][y]) {if (x!=1 &&!map[xx] [YY]) FloodFill (XX,YY);} else if (map[xx][yy]) Add (T,map[xx][yy]), add (map[xx][yy],t), Xx=x+1,yy=y;if (!map2[x+1][y]) {if (x!=m &&!map[xx][yy]) FloodFill ( XX,YY);} else if (Map[xx][yy]) Add (T,map[xx][yy]), add (map[xx][yy],t);} void BFS () {static int q[65540];static unsigned short r,h;int i;f[q[++r]=map[xa][ya]]=1;while (r!=h) {int x=q[++h];for (i= Head[x];i;i=table[i].next) if (!f[table[i].to]) f[q[++r]=table[i].to]=f[x]+1;}} int main () {int i,j;cin>>cnt;for (i=1;i<=cnt;i++) cin>>a[i];cin>>xa>>ya>>xb> >yb;for (i=1;i<=cnt;i++) {B[i+i-1]=make_pair (a[i].x,&rectangles[i].x1); B[i<<1]=make_pair (a[i].x+ A[I].C,&AMP;RECTANGLES[I].X2);} B[cnt<<1|1]=make_pair (XA,&AMP;XA); B[cnt+1<<1]=make_pair (XB,&AMP;XB); sort (b+1,b+ (cnt+1<<1) +1) ; for (i=1;i<= (cnt+1<<1); i++) {if (i==1| | B[i].first!=b[i-1].first) ++m;*b[i].second=m;} M+=2;for (i=1;i<=cnt;i++) {B[i+i-1]=make_pair (a[i].y,&rectangles[i].y1); B[i<<1]=make_pair (a[i].y+a[ I].C,&AMP;RECTANGLES[I].Y2);} B[cnt<<1|1]=make_pair (YA,&AMp;ya); B[cnt+1<<1]=make_pair (YB,&AMP;YB); sort (b+1,b+ (cnt+1<<1) +1); for (i=1;i<= (cnt+1<<1); I + +) {if (i==1| | B[i].first!=b[i-1].first) ++n;*b[i].second=n;} N+=2;for (i=1;i<=cnt;i++) {for (j=rectangles[i].x1;j<rectangles[i].x2;j++) map1[j][rectangles[i].y1]=map1[j] [Rectangles[i].y2]=true;for (j=rectangles[i].y1;j<rectangles[i].y2;j++) map2[rectangles[i].x1][j]=map2[ Rectangles[i].x2][j]=true;} for (i=1;i<=m;i++) for (j=1;j<=n;j++) if (!map[i][j]) ++t,floodfill (I,J); BFS (); Cout<<f[map[xb][yb]]-1<<endl;return 0;}

Bzoj 1967 Ahoi2005 Cross through magnetic field FLOODFILL+BFS