# sgu176 Flow Construction "There are Yuanhui with the lowest flow in the upper and lower bounds"

Source: Internet
Author: User

The same is the template problem.

First will have Yuanhui conversion to no Yuanhui, assuming the original Yuanhui for St, we join the Yuanhui for St, then we should from T to s even a flow for the +∞ edge, so that the original St to meet the balance of payments, degenerate into a common node.

Separating the necessary and other edges, from S to T run the maximum flow, all the edges connected to the source or sink are full and prove to be solvable.

Remove the t to s capacity for the +∞ edge, remove the necessary edges, and run the maximum stream from T to S.

If the answer we get is negative, then it means that the internal T to S is connected to the ring, then we add s to s capacity for the-ans edge, run s to t maximum flow, so that all sides of the traffic should be 0, plus the lower bound of traffic is the answer.

`1#include <bits/stdc++.h>2 #defineRep (I, A, b) for (int i = A; I <= b; i++)3 #defineDrep (I, A, b) for (int i = A; I >= b; i--)4 #defineREP (I, A, b) for (int i = A; I < b; i++)5 #defineMP Make_pair6 #definePB Push_back7 #defineCLR (x) memset (x, 0, sizeof (x))8 #defineXX First9 #defineyy secondTen using namespacestd; OnetypedefLong Longi64; Atypedef pair<int,int>PII; - Const intINF = ~0U>>1; - Consti64 INF = ~0ull >>1; the //****************************** -  - Const intMAXN = the, MAXM =10105; -  + structEd { -     intu, V, NX, C; Ed () {} +Ed (int_u,int_v,int_NX,int_c): A u (_u), V (_v), NX (_NX), C (_c) {} at} E[MAXM <<1]; - intG[MAXN], Edtot =1; - voidAddedge (intUintVintc) { -E[++edtot] =Ed (U, V, g[u], c); -G[u] =Edtot; -E[++edtot] = Ed (V, u, G[v],0); inG[V] =Edtot; - } to  + intLEVEL[MAXN], S, T; - BOOLBFsintSintt) { the     Static intQUE[MAXN];intQh0), QT (0); *CLR (level); LEVEL[QUE[++QT] = s] =1; \$      while(QH! =qt) {Panax Notoginseng         intx = Que[++qh];if(QH = = MAXN-1) QH =0; -          for(inti = g[x]; I i = e[i].nx)if(E[I].C &&!)LEVEL[E[I].V]) { theLEVEL[QUE[++QT] = e[i].v] = level[x] +1; +             if(qt = = MAXN-1) qt =0; A         } the     } +     return!!Level[t]; - } \$ intDfsintUintRmintt) { \$     if(U = = t)returnRM; -     intRM1 =RM; -      for(inti = G[u]; I i =e[i].nx) { the         if(e[i].c && level[e[i].v] = = Level[u] +1) { -             intFlow =dfs (E[I].V, min (e[i].c, RM), t);WuyiE[I].C-= flow, E[i ^1].C + =flow; the             if(RM-= flow) = =0) Break; -         } Wu     } -     if(Rm1 = = RM) Level[u] =0; About     returnRM1-RM; \$ } -  - intL[MAXM],inch[MAXN]; - intMain () { A     intN, M; +scanf"%d%d", &n, &m); theRep (I,1, M) { -         intX, Y, a, B; scanf"%d%d%d%d", &x, &y, &a, &b); \$L[i] = A *b; the         if(b)inch[y] + = A,inch[x]-=A; theAddedge (x, y, a-a *b); the     } theS = n +1, T = n +2; -     intSum0); inRep (I,1, N) { the         if(inch[I] >0) Sum + =inch[i]; the         if(inch[I] >0) Addedge (S, I,inch[i]); About         ElseAddedge (i, T,-inch[i]); the     } theAddedge (N,1,0x3f3f3f3f); the      while(BFS (S, T)) Sum-= DFS (s,0x3f3f3f3f, T); +     if(sum) {Puts ("Impossible");return 0; } -     intAns =e[edtot].c; thee[edtot].c = E[edtot ^1].C =0;Bayi     inttmp0); the      while(BFS (N,1)) tmp + = DFS (n,0x3f3f3f3f,1); theAns-=tmp; -     if(Ans <0) { -Addedge (S),1, -ans); the          while(BFS (S, N)) Dfs (s,0x3f3f3f3f, n); theAns =0; the     } theprintf"%d\n", ans); -Rep (I,1, M) { theprintf (i = =1?"%d":"%d", e[(i <<1) ^1].C +l[i]); the     } thePuts"");94     return 0; the}`
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sgu176 Flow Construction "There are Yuanhui with the lowest flow in the upper and lower bounds"

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