Source: Internet
Author: User

For a diagram, some points need to be connected to 1 points individually in a certain color line, asking how to arrange the colors that make the whole graph the most color of the road with the least color.

Obviously, the binary enumeration and then color is actually the flow, equivalent to each edge limit a current two points of traffic value, determine whether the full stream can be.

Summon Code June:

`#include <iostream>#include<cstdio>#include<cstring>#defineMAXN 555#defineMAXM 333333using namespacestd;Const intinf=~0U>>2;intTo[maxm],next[maxm],c[maxm],first[maxm],edge;inttag[maxn],d[maxn],tag=520;intQ[maxn],bot,top;BOOLCAN[MAXN];intN,m,k,t,h,s,t,house,ans,boder;voidAddedge (intUintV) {Edge++; To[edge]=v,c[edge]=1, next[edge]=first[u],first[u]=Edge; Edge++; To[edge]=u,c[edge]=1, next[edge]=first[v],first[v]=Edge;}void_input () {scanf ("%d%d%d",&n,&m,&k); Edge=-1;  for(intI=0; i<=n; i++) first[i]=-1; S=1, t=0, house=K;  while(k--) {scanf ("%d",&h);    Addedge (h,t); } boder=Edge;  while(m--) {scanf ("%d%d",&k,&h);    Addedge (K,H); }}BOOLBFs () {Q[bot=top=1]=t,d[t]=0, can[t]=false, tag[t]=++TAG;  while(bot<=top) {        intcur=q[bot++];  for(intI=first[cur]; i!=-1; I=Next[i]) {            if(c[i^1]>0&& tag[to[i]]!=TAG) {Tag[to[i]]=tag,q[++top]=To[i]; D[to[i]]=d[cur]+1, can[to[i]]=false; if(to[i]==s)return true; }        }    }    return false;}intDfsintCurintnum) {    if(cur==t)returnnum; inttmp=num,k;  for(intI=first[cur]; i!=-1; I=Next[i])if(c[i]>0&& d[to[i]]==d[cur]-1&& Tag[to[i]]==tag &&!Can[to[i]]) {k=Dfs (To[i],min (num,c[i)); if(k) num-=k,c[i]-=k,c[i^1]+=K; if(num==0) Break; }    if(num) can[cur]=true; returntmp-num;}BOOLCheckintx) {     for(intI=0; i<=boder; i++) c[i]=1;  for(inti=boder+1; i<=edge; i++) c[i]=x;  for(ans=0; BFS (); ) ans+=DFS (S,inf); returnans>=House ;}intMain () {scanf ("%d",&T);  while(t--) {_input (); intL=1, r=N,mid;  while(l<r) {Mid= (l+r) >>1; if(Check (mid)) r=mid; ElseL=mid+1; } printf ("%d\n", L); }    return 0;}`
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