Network flow + two points.
The N^3 enumeration determines the sprites that each lich can attack, connecting 1 of its edges, each of which has a 1 edge to the sink point.
Two-point answer, modify the source points to each lich's cap, run the maximum flow to see if it equals the number of sprites.
Well, there's nothing wrong with it.
And then mad WA not only. For one night. Most of the night, found on the internet to get the program is a WA ... What else can I say?
It was only then that I should count the distance to the line rather than the line. Keep smiling.
So there is a tag for calculating the geometry of this problem?
Calculate the distance from point to line D, point to line two points distance, short for L, long for R.
The Pythagorean theorem calculates tmp=sqrt (r*r-d*d); if TMP is less than the segment length, then D is returned, otherwise l;
//Twenty#include <cstdio>#include<cstdlib>#include<iostream>#include<algorithm>#include<cmath>#include<cstring>#include<queue>#include<vector>#defineINF 0x3f3f3f3ftypedefLong LongLL;Const intmaxn=299*299*2+5;using namespacestd;intAns,s,t,n,u,v,w,ecnt=1, FIR[MAXN],D[MAXN],CUR[MAXN],C[MAXN],P[MAXN],ED[MAXN];structwuyao{intx,y,r,t;} WY[MAXN],JL[MAXN],SM[MAXN];structEdge {int from, TO,CAP,FLOW,NXT; Edge () {} Edge (int from,intTo,intCapintFlowintNXT): from( from), to, Cap (CAP), flow (flow), NXT (NXT) {}}E[MAXN];voidAddintUintVintW) {e[++ecnt]=edge (U,v,w,0, Fir[u]); e[++ecnt]=edge (V,u,0,0, Fir[v]); Fir[u]=ecnt-1; fir[v]=ecnt;} Queue<int>que;voidBFsintSintt) { for(intI=1; i<=n;i++) d[i]=N; D[t]=0; Que.push (t); while(!Que.empty ()) { intx=Que.front (); Que.pop (); for(intI=fir[x];i;i=e[i].nxt)if(d[e[i].to]==n&&e[i].flow==E[i].cap) {D[e[i].to]=d[x]+1; Que.push (e[i].to); } }}intCalintSintt) {intFl=INF; for(intX=T;X!=S;X=E[P[X]]. from) FL=min (fl,e[p[x]].cap-E[p[x]].flow); for(intX=T;X!=S;X=E[P[X]]. from) {E[p[x]].flow+=FL; E[P[X]^1].flow-=FL; } returnFL;}intMaxflow (intSintt) {BFS (s,t); intres=0; for(intI=1; i<=n;i++) cur[i]=fir[i],c[d[i]]++; for(intx=s;d[x]<N;) { if(x==t) {res+=cal (S,t); X=s; } intok=0; for(int&i=cur[x];i;i=e[i].nxt)if(d[e[i].to]+1==d[x]&&e[i].cap>E[i].flow) {p[x=e[i].to]=i; OK=1; Break; } if(!OK) {Cur[x]=FIR[X];intm=N; for(intI=cur[x];i;i=e[i].nxt)if(e[i].cap>E[i].flow) M=min (m,d[e[i].to]+1); if(! (--c[d[x])) Break; C[D[X]=m]++; if(X!=s) x=e[p[x]]. from; } } returnRes;}DoubleDisintXintYintXxintyy) { returnsqrt ((Double) (X-XX) * (X-XX) + (Double) (y-yy) * (Y-yy));}DoubleDis (intXintYDoubleADoubleBDoubleC) {returnFabs ((a*x+b*y+c))/sqrt (a*a+b*B);} intwys,jls,sms;DoubleYyj;intCheckintti) { for(intI=1; i<=ecnt;i++) e[i].flow=0; for(intI=1; i<=wys;i++) { intfl=ti/wy[i].t+1; E[ed[i]].cap=FL; } return(Maxflow (s,t) = =JLS);}intMain () {//freopen (". In", "R", stdin); //freopen (". Out", "w", stdout);scanf ("%d%d%d",&wys,&jls,&SMS); N=wys+jls+2; s=n-1; t=N; for(intI=1; i<=wys;i++) {scanf ("%d%d%d%d",&wy[i].x,&wy[i].y,&wy[i].r,&wy[i].t); Add (S,i,0); ed[i]=ecnt-1; } for(intI=1; i<=jls;i++) {scanf ("%d%d",&jl[i].x,&jl[i].y); Add (Wys+i,t,1); } for(intI=1; i<=sms;i++) scanf ("%d%d%d",&sm[i].x,&sm[i].y,&SM[I].R); for(intI=1; i<=wys;i++) { for(intj=1; j<=jls;j++) if((Yyj=dis (WY[I].X,WY[I].Y,JL[J].X,JL[J].Y)) <= (Double) {WY[I].R) {DoubleA= (WY[I].Y-JL[J].Y), b=jl[j].x-wy[i].x,c=wy[i].x*jl[j].y-wy[i].y*jl[j].x; if(i== -) { intdebug=1; } if(!SMS) Add (I,wys+j,1); for(intk=1; k<=sms;k++) { Doubletmp; Doubleddx=Dis (SM[K].X,SM[K].Y,A,B,C); DoubleZb=dis (WY[I].X,WY[I].Y,SM[K].X,SM[K].Y), yb=dis (JL[J].X,JL[J].Y,SM[K].X,SM[K].Y); if(zb<yb) Swap (ZB,YB); DoubleWOC=SQRT (zb*zb-ddx*DDX); TMP=woc<=yyj?Ddx:yb; if(Tmp<= (Double) SM[K].R) Break; if(k==SMS) Add (I,wys+j,1); } } } intL=0, r=4e6+5; if(!check (R)) ans=-1; Else { while(l<=r) {intMid= (l+r) >>1; if(Check (mid)) ans=mid,r=mid-1; ElseL=mid+1; }} printf ("%d\n", ans); return 0;}
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Bzoj 1822[jsoi2010]frozen Nova Frozen Wave