Poj 3801 HDU 3157 crazy circuits-a source sink has the upper and lower bounds of the smallest stream

/* HDU 3157poj 3801 question: a circuit board with N wiring bars (Number 1 ~ N) there are also two power supply terminals +-and then the terminals and the minimum current of M parts are given to find a total current that can make all parts work normally. Then the output impossible is actually a lower bound with a source sink. minimum flow problem handling: 1. construct an additional network 2. find the maximum stream for SS and TT (the full stream of SS and TT has a solution) 3. if there is a solution, find the maximum stream for S and T, and the source sink has the upper and lower bounded minimum flow problem is: 1. create an additional network (without adding [t, s] edges) 2. find the maximum stream for SS and TT 3. add the [t, s] side 4. find the maximum stream for SS and TT 5. if the SS and TT streams are full, the [t, s] traffic is the smallest stream. Code Most of them are extracted from other questions, and comments are messy */# include <stdio. h> # include <string. h> # define INF 0x7fffffffstruct edge // edge {int U, V, F, next, B, c; // Number of the bottom edge of the stream at the front node and the back node of the edge} e [1500]; int head [70], in [70], S, t, SS, TT, Yong, sum; int n, m; void INI () {memset (Head,-1, sizeof (head); Yong = 0; memset (in, 0, sizeof (in); s = 0, T = n + 1, SS = t + 1, TT = SS + 1; // sum = 0;} void Adde (int from, int to, int Xia, int Shang) // Add edge {// Add edge e [Yong]. U = from, E [Yong]. V =, E [Yong]. F = Shang-xia, E [Yong]. B = Xia, E [Yong]. C = Shang; E [Yong]. next = head [from], head [from] = Yong ++; // Add its return edge e [Yong] at the same time. U = To, E [Yong]. V = from, E [Yong]. f = 0, E [Yong]. B = Xia, E [Yong]. C = Shang; E [Yong]. next = head [to], head [to] = Yong ++;} void build () {int I; for (I = 0; I <= T; ++ I) {If (in [I]> 0) Adde (SS, I, 0, in [I]); else {Adde (I, TT, 0, -In [I]); sum + = (-in [I]) ;}} int d [1000], num [1000]; int min (int, int B) {return a <B? A: B;} int sap_gap (int u, int F, int S, int t) // recursive sap {If (u = T) return F; int I, V, mind = T, last = F, cost; for (I = head [u]; I! =-1; I = E [I]. next) {v = E [I]. v; int flow = E [I]. f; If (flow> 0) // when writing a reference template, the flow is written as f {If (d [u] = d [v] + 1) {cost = sap_gap (v, min (last, flow), S, T); e [I]. f-= cost; E [I ^ 1]. F + = cost; last-= cost; If (d [s]> = t + 1) return F-last; If (last = 0) break ;} if (d [v] <mind) Mind = d [v] ;}} if (last = f) {-- num [d [u]; if (Num [d [u] = 0) d [s] = t + 1; d [u] = mind + 1; ++ num [d [u];} return F-last;} int max_f (int s, int t) {int f = 0; memset (D, 0, size Of (d); memset (Num, 0, sizeof (Num); For (Num [s] = t + 1; d [s] <t + 1 ;) F + = sap_gap (S, INF, S, T); Return F;} int main () {int I, dat, U, V, F1, F2, P; char from [10], to [10]; while (scanf ("% d", & N, & M), N + M) {INI (); for (I = 1; I <= m; ++ I) {scanf ("% S % d", from, to, & dat ); if (from [0] = '+') u = s; else sscanf (from, "% d", & U ); if (to [0] = '-') V = T; else sscanf (to, "% d", & V); Adde (u, v, dat, INF); in [u]-= dat, in [v] + = dat;} build (); F1 = max_f (SS, TT); P = Yong; Adde (T, S, 0, INF); F2 = max_f (SS, TT); If (F1 + F2! = Sum) printf ("impossible \ n"); else printf ("% d \ n", E [P ^ 1]. f);} return 0 ;}