Freud Floyd to find the smallest ring

Template :

#include <bits/stdc++.h>using namespacestd;Const intMAXN = the;Const intINF =0xffffff0;inttemp,map[maxn][maxn],dist[maxn][maxn],pre[maxn][maxn],ans[maxn*3];voidSolve (intIintJintk) {Temp=0;//backtracking, storage of the smallest ring     while(I! =j) {Ans[temp++] =J; J=Pre[i][j]; } ans[temp++] =i; Ans[temp++] =K;}voidFloyd (intN) {     for(inti =1; I <= N; ++i) for(intj =1; J <= N; ++j) {Dist[i][j]=Map[i][j]; PRE[I][J]=i; }    intMincircle = INF;//Minimum Ring     for(intK =1; K <= N; ++k) { for(inti =1; I <= N; ++i) { for(intj =1; J <= N; ++j) {if(I! = J && Dist[i][j]! = INF && map[i][k]! = INF && map[k][j]! =INF&& Dist[i][j] + map[i][k] + Map[k][j] <mincircle) {mincircle= Min (mincircle, dist[i][j] + map[i][k] +Map[k][j]);   Solve (I,J,K); //Back storage Minimum ring                }            }        }         for(inti =1; I <= N; ++i) { for(intj =1; J <= N; ++j) {if(Dist[i][k]! = INF && dist[k][j]! = INF &&Dist[i][k]+ Dist[k][j] <Dist[i][j]) {Dist[i][j]= Dist[i][k] +Dist[k][j]; PRE[I][J]= Pre[k][j];//Record point I point J on the path, J Front Dot                }            }        }    }    if(mincircle = = INF)//There is no ring{printf ("No solution.\n"); return; }    //If the minimum ring is found to be negative, the original image must have a negative ring     for(inti =0; i < temp; ++i)//Output Minimum ring        if(I! = temp-1) printf ("%d", Ans[i]); Elseprintf ("%d\n", Ans[i]);}

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Example :

BZOJ1027: [JSOI2007] alloy

Idea: given two point setsAAndBPleaseAA subset of the smallestSMakeBAll the points in theSInside the convex package. EnumerationAPoint set two pointsI,J(Ican be equal toJ) ifBAll points in the point set are in the vectori->jThe left or segment of theIJ, connect ai->jOne-way edges, and thenFloydThe smallest ring can be obtained.

#include <bits/stdc++.h>#defineEPS 1e-8using namespacestd;structnode{Doublex, Y, Z;} a[510],b[510];Doublemulti (node P1,node p2,node p0) {Doublex1=p1.x-p0.x; Doubley1=p1.y-p0.y; Doublex2=p2.x-p0.x; Doubley2=p2.y-p0.y; returnx1*y2-x2*Y1;}DoubleDis (node p1,node p2) {returnsqrt ((p1.x-p2.x) * (p1.x-p2.x) + (P1.Y-P2.Y) * (p1.y-p2.y));}intd[510][510];intMain () {intn,m; scanf ("%d%d",&m,&N);  for(intI=1; i<=m;i++) scanf ("%LF%LF%LF",&a[i].x,&a[i].y,&a[i].z);  for(intI=1; i<=n;i++) scanf ("%LF%LF%LF",&b[i].x,&b[i].y,&b[i].z); intI,j,k;DoubleT; memset (d,0x3f,sizeof(d));  for(i=1; i<=m;i++)         for(j=1; j<=m;j++)        {             for(k=1; k<=n;k++) {T=multi (a[j],b[k],a[i]); if(t<-eps) Break; if(Fabs (t) <eps && Dis (a[i],b[k]) >dis (A[i],a[j])) Break; }            if(k==n+1) d[i][j]=1; }     for(k=1; k<=m;k++)         for(i=1; i<=m;i++)if(i!=k) for(j=1; j<=m;j++)if(j!=k)if(D[i][j]>d[i][k]+d[k][j]) d[i][j]=d[i][k]+D[k][j]; intans=m+1;  for(i=1; i<=m;i++)if(Ans>d[i][i]) ans=D[i][i]; if(ans==m+1) printf ("-1\n");Elseprintf"%d\n", ans); return 0;}

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Freud Floyd to find the smallest ring