BZOJ1018: [SHOI2008] clogged traffic traffic

Segment tree maintains connectivity.

Below the topic there is analysis.

The number of lines in the puzzle is equal to the number of rows in the title-1.

analysed altogether (R1,C1)--(R2,C2) a total of four ways.

Suppose R1=1,r2=0.

1. Go directly to the past.

2. First come (0,C1) and then go over.

3. First come (1,C2) and then go over.

4. Go first (0,C1) and then (1,C2) to the past.

With an array of a1[i][j], maintain the line of I from the leftmost X-node to the far right of J-line.

Another array, a2[i],a2[0], indicates that the point at the far right of the X node can be wrapped from one side to another. A2[1] represents the leftmost point on the left to be able to move from the left to another line.

#include <cstdio>#include<algorithm>#include<cstring>using namespacestd;Const intMAXN =800000+Ten;Chars[ -];intx1,x2,y1,y2,c;structSegtree {#defineLC (x) ((x) <<1)#defineRC (x) (((x) <<1) |#defineRoot 1structStatus {inta1[2][2],a2[2];        }S[MAXN]; BOOLb[maxn][2]; intL[MAXN],R[MAXN],M[MAXN]; Status update (Status S1,status S2,BOOLb[])        {Status res;  for(intI=0; i<=1; i++)          for(intj=0; j<=1; j + +) Res.a1[i][j]=s1.a1[i][0] && b[0] && s2.a1[0][J] | | s1.a1[i][1] && b[1] && s2.a1[1][j]; res.a2[0]=s1.a2[0] || s1.a1[0][0] && b[0] && s2.a2[0] && b[1] && s1.a1[1][1]; res.a2[1]=s2.a2[1] || s2.a1[0][0] && b[0] && s1.a2[1] && b[1] && s2.a1[1][1]; returnRes; } Status Access (intXintY1,inty2) {        if(Y1<=l[x] && r[x]<=y2)returnS[x]; Else if(Y2<=m[x])returnAccess (LC (x), y1,y2); Else if(Y1>m[x])returnAccess (RC (x), y1,y2); Else returnUpdate (Access (LC (x), y1,y2), Access (RC (x), y1,y2), b[x]); }        voidChangeBOOLKintXintX1,intY1,intX2,inty2) {        if(X1==x2 && y1==M[x]) {B[X][X1]=K; S[X]=Update (S[LC (x)],S[RC (x)],b[x]); }        Else if(l[x]==r[x]) s[x].a1[0][1]=s[x].a1[1][0]=s[x].a2[0]=s[x].a2[1]=K; Else{Change (k,y2&LT;=M[X]?LC (X): RC (x), x1,y1,x2,y2); S[X]=Update (S[LC (x)],S[RC (x)],b[x]); }    }        voidAskintX1,intY1,intX2,inty2) {Status Left=access (Root,1, y1), right=Access (root,y2,c), Mid=access (ROOT,Y1,Y2); BOOLres=false;  for(intI=0; i<=1; i++)         for(intj=0; j<=1; j + +)            if(mid.a1[i][j]&& (i==x1| | left.a2[1]) && (j==x2| | right.a2[0]) {res=true; Break; } printf (res?"y\n":"n\n"); }        voidBuildintXintY1,inty2) {L[x]=y1,r[x]=y2,m[x]= (L[x]+r[x]) >>1; if(y1==y2) s[x].a1[0][0]=s[x].a1[1][1]=true; Else{Build (LC (x), y1,m[x]); Build (RC (x), m[x]+1, y2); }}}seg;intMain () {scanf ("%d",&c); Seg.build (1,1, c);  while(1) {scanf ("%s", s); if(s[0]=='E') Break; scanf ("%d%d%d%d",&x1,&y1,&x2,&y2); --x1; --x2; if(y1>y2)            {swap (X1,X2);            Swap (Y1,Y2); }        if(s[0]=='O') Seg.change (1,1, X1,y1,x2,y2); Else if(s[0]=='C') Seg.change (0,1, X1,y1,x2,y2); ElseSeg.ask (X1,Y1,X2,Y2); }        return 0; }

BZOJ1018: [SHOI2008] clogged traffic traffic