bzoj1061: [Noi2008] Volunteer recruitment

Cost flow.

The legend of the modeling of the synthesis of the question ...

Https://www.byvoid.com/blog/noi-2008-employee

There is a whole lot of talk about modeling.

I write and write the actual meaning of modeling according to my own understanding.

If a[i]-a[i-1]>=0, the representative needs new hires. Otherwise, it means someone has left. (Leave does not have to be in t[i]+1 days, if the latter person is superfluous, will leave very early).

1. The model is s->i capacity of a[i]-a[i-1], the cost is 0. 2.i->t capacity is – (A[i]-a[i-1)) at a cost of 0.

For each person he will enter in S[i] days, and leave at (t[i]+1) day or earlier.

3. In s[i] to t[i]+1 a capacity unlimited cost for the side of C[i]. 4. On the side of (i+1)->i a capacity unlimited cost of 0.

s represents the person entering, and T represents the person who left.

The most important is the idea of flow balance when modeling.

Set Y[i] for the extra person for the first day. X[i] is the number of volunteers of type I.

For each point (A[i]-a[i-1]) + (y[i]-y[i-1]) +sum (X[k1])-sum (X[k2]) =0 (K1 and K2 meet t[k1]+1=i,s[k2]=i).

Each point satisfies such a flow balance

#include <cstdio>#include<algorithm>#include<cstring>using namespaceStd;typedefLong LongLL;Const intMAXN = ++Ten;Const intMAXM =100000+Ten;Const intINF =0x3f3f3f3f;intG[maxn],v[maxm],f[maxm],c[maxm],next[maxm],eid;intn,m,s,t;intA[MAXN];Long LongRes; LL DIST[MAXN];intPRE[MAXN];BOOLINQUE[MAXN];intQ[maxm],l,r,u;voidAddedge (intAintBintFintC) {V[eid]=b; F[eid]=f; C[eid]=c; Next[eid]=g[a]; g[a]=eid++; V[eid]=a; f[eid]=0; C[eid]=-c; NEXT[EID]=G[B]; g[b]=eid++; }voidbuild () {memset (g,-1,sizeof(g)); scanf ("%d%d",&n,&m); S=0; t=n+2;  for(intI=1; i<=n;i++) scanf ("%d",&A[i]);  for(intI=1, d;i<=n+1; i++) {D=a[i]-a[i-1]; if(d>=0) Addedge (S,i,d,0); ElseAddedge (i,t,-d,0); }     for(intI=1, s,t,c;i<=m;i++) {scanf ("%d%d%d",&s,&t,&B); Addedge (S,t+1, Inf,c); }     for(intI=1; i<=n;i++) Addedge (i+1, I,inf,0);}BOOLSPFA () {memset (dist,0x3f,sizeof(Dist)); L=r=0; Q[r++]=S; Dist[s]=0;  while(l<r) {Inque[u=q[l++]]=0;  for(intI=g[u];~i;i=next[i])if(F[i] && dist[v[i]]>dist[u]+C[i]) {Dist[v[i]]=dist[u]+C[i]; Pre[v[i]]=i; if(!inque[v[i]]) inque[q[r++]=v[i]]=1; }    }    returndist[t]<inf;}Long Longaugment () {Long Longaug=inf,res=0;  for(inti=t;i!=s;i=v[pre[i]^1]) Aug=min (Long Long) f[pre[i]);  for(inti=t;i!=s;i=v[pre[i]^1]) {F[pre[i]]-=; F[pre[i]^1]+=; Res+=aug*C[pre[i]]; }    returnRes;}voidsolve () {Long Longres=0;  while(SPFA ()) res+=augment (); printf ("%lld\n", res);}intMain () {build ();    Solve (); return 0;}

bzoj1061: [Noi2008] Volunteer recruitment