Codeforces 653D Delivery Bears (maximum flow)

The topic probably said that there is a simple graph of n-point M-side, each side can only allow a certain amount of goods through. To allow the X-Bear to deliver goods from 1 to n, each bear is transported and transported in the same weight, asking for the maximum value of the weight.

The two-part weight judgment is established.

If the weight is known, then how many bears can be found on each side, that is, the capacity of the side divided by the weight. And judging whether the X-Bear can go to 1 to N is determined by the maximum flow.

Then the problem is OK--but pay attention to the accuracy--there will be a burst int!. I really can't believe it.

1#include <cstdio>2#include <cstring>3#include <cmath>4#include <queue>5#include <algorithm>6 using namespacestd;7 #defineINF (1&LT;&LT;30)8 #defineMAXN 55559 #defineMAXM 11111Ten  One structedge{ A     intV,cap,flow,next; - }EDGE[MAXM]; - intVs,vt,ne,nv; the intHEAD[MAXN]; -  - voidAddedge (intUintVintcap) { -Edge[ne].v=v; Edge[ne].cap=cap; edge[ne].flow=0; +Edge[ne].next=head[u]; head[u]=ne++; -Edge[ne].v=u; edge[ne].cap=0; edge[ne].flow=0; +EDGE[NE].NEXT=HEAD[V]; head[v]=ne++; A } at  - intLEVEL[MAXN]; - intGAP[MAXN]; - voidBFs () { -memset (level,-1,sizeof(level)); -Memset (Gap,0,sizeof(GAP)); inlevel[vt]=0; -gap[level[vt]]++; toqueue<int>que; + Que.push (VT); -      while(!Que.empty ()) { the         intu=Que.front (); Que.pop (); *          for(intI=head[u]; i!=-1; I=Edge[i].next) { $             intv=edge[i].v;Panax Notoginseng             if(level[v]!=-1)Continue; -level[v]=level[u]+1; thegap[level[v]]++; + Que.push (v); A         } the     } + } -  $ intPRE[MAXN]; $ intCUR[MAXN]; - intIsap () { - BFS (); thememset (pre,-1,sizeof(pre)); -memcpy (Cur,head,sizeof(head));Wuyi     intu=pre[vs]=vs,flow=0, aug=INF; thegap[0]=NV; -      while(level[vs]<NV) { Wu         BOOLflag=false; -          for(int&i=cur[u]; i!=-1; I=Edge[i].next) { About             intv=edge[i].v; $             if(Edge[i].cap!=edge[i].flow && level[u]==level[v]+1){ -flag=true; -pre[v]=u; -u=v; A                 //aug= (Aug==-1?edge[i].cap:min (Aug,edge[i].cap)); +Aug=min (aug,edge[i].cap-edge[i].flow); the                 if(v==VT) { -flow+=; $                      for(u=pre[v]; V!=vs; v=u,u=Pre[u]) { theedge[cur[u]].flow+=; theedge[cur[u]^1].flow-=; the                     } the                     //Aug=-1; -aug=INF; in                 } the                  Break; the             } About         } the         if(flag)Continue; the         intMinlevel=NV; the          for(intI=head[u]; i!=-1; I=Edge[i].next) { +             intv=edge[i].v; -             if(Edge[i].cap!=edge[i].flow && level[v]<minlevel) { theMinlevel=Level[v];Bayicur[u]=i; the             } the         } -         if(--gap[level[u]]==0) Break; -level[u]=minlevel+1; thegap[level[u]]++; theu=Pre[u]; the     } the     returnflow; - } the  the int  from[MAXN],TO[MAXN],CAP[MAXN]; the intn,m,x;94 BOOLisOKDoublemid) { theXss0; Vt=n; nv=n+1; Ne=0; thememset (head,-1,sizeof(head)); theAddedge (VS,1, x);98      for(intI=0; i<m; ++i) { About         Long LongTmp= (Long Long) (cap[i]/mid+1e-9); -         if(Tmp>inf) tmp=INF;101Addedge ( from[i],to[i],tmp];102     }103     returnISAP () = =x;104 } the 106 intMain () {107scanf"%d%d%d",&n,&m,&x);108      for(intI=0; i<m; ++i) {109scanf"%d%d%d", from+i,to+i,cap+i); the     }111     DoubleL=0, r=1000000, mid; the      while(Fabs (r-l) >1e-9){113Mid= (L+R)/2; the         if(mid==0) Break; the         if(isOK (mid)) { theL=mid;117}Else{118R=mid;119         } -     }121printf"%f\n", mid*x);122     return 0;123}

Codeforces 653D Delivery Bears (maximum flow)