UVA-12264 Risk

Test instructions comparison pit, after moving the soldiers can not move again, or the sample will not be over ...

Minimum value maximum so the two-point answer, run network flow verification is feasible.

This topic focus on the building, in order to ensure that only move once, split, one into the point of an out point, to the point of nature can not be dispatched again.

An edge that does not have a capacity of 1 on the edge of the boundary means that at least one person is left, and the edge of the T on the border is set to mid.

Other Details:

Pay attention to the two points when the answer is L=mid+1,r=mid (l+r)/2 downward rounding,

L=mid,r=mid-1 (l+r)/2 rounding up, otherwise encountered such as R = l+1 judgment after the execution L = Mid this situation will die loop.

At first I wanted to reduce the flow of traffic no longer on the boundary by one, but it was quite similar to forcing the current position toward T by 1, in fact its 1 did not have to be provided by itself

, after moving the soldiers can not move again, then the flow can be reduced to the border, unless the movement after the soldiers can move again this is the right. Test Instructions Pit Ah ...

#include <bits/stdc++.h>using namespacestd;Const intMAXN = the<<1;intA[maxn],n;structedge{intv,cap,flow,nxt;}; Vector<Edge>edges;#definePB push_backintHEAD[MAXN];voidAddedge (intUintVintc) {edges. PB ({v,c,0, Head[u]}); Head[u]= Edges.size ()-1; Edges. PB ({u,0,0, Head[v]}); HEAD[V]= Edges.size ()-1;}Const intINF =0x3f3f3f3f;intS,T,CUR[MAXN],Q[MAXN],D[MAXN];BOOLBFs () {memset (d,0,sizeof(d)); intL =0, r =0; Q[r+ +] = S; D[s] =1;  while(r>l) {        intU = q[l++];  for(inti = Head[u]; ~i; i =edges[i].nxt) {Edge&e =Edges[i]; if(!D[E.V] && e.cap>E.flow) {D[E.V]= d[u]+1; Q[r++] =e.v; }        }    }    returnd[t];}intDfsintUinta) {    if(U = = t| |! AreturnA; intFlow =0, F;  for(int&i = Cur[u]; ~i; i =edges[i].nxt) {Edge&e =Edges[i]; if(D[E.V] = = d[u]+1&& (f = DFS (E.v,min (e.cap-e.flow,a))))  {Flow+ = f; A-=F; E.flow+ = f; edges[i^1].flow-=F; if(!a) Break; }    }    returnflow;}intMaxflow () {intFlow =0;  while(BFS ()) {memcpy (Cur,head,sizeof(head)); Flow+=DFS (S,inf); }    returnflow;} Vector<int>Change ;voidinit () {edges.clear ();    Change.clear (); memset (Head,-1,sizeof(head));}voidRebuildintcap) {     for(inti =0; I < edges.size (); i++) {Edges[i].flow=0; }     for(inti =0; I < change.size (); i++) {Edges[change[i]].cap=cap; }}intMain () {//freopen ("In.txt", "R", stdin);    intTest; scanf"%d",&Test);  while(test--) {scanf ("%d",&N);        Init (); S= n<<1; T = s|1; intL =1, r =0;  for(inti =0; I < n; i++) {scanf ("%d", a+i); R + =A[i]; }        CharSTR[MAXN]; intCNT =0, sum =0;  for(inti =0; I < n; i++) {scanf ("%s", str); if(!a[i])Continue; BOOLborder =false;  for(intj =0; J < N; J + +){                if(Str[j] = ='Y'){                    if(A[j]) Addedge (i,j+N,inf); ElseBorder=true;            }} Addedge (S,i,a[i]); Addedge (I,i+N,inf); if(border) CHANGE.PB (Edges.size ()), Addedge (I+n,t,0), cnt++; ElseAddedge (I+n,t,1), sum++; }        intmid;  for(; l < R; sum + mid*cnt = = Maxflow ()? l = Mid:r = mid-1) Mid= (l+r+1) >>1, rebuild (mid);//L = Mid. Rounding up L = mid+1 roundingprintf"%d\n", L); }    return 0;}

UVA-12264 Risk