Sgu 176 upstream and downstream minimum streams

There are m pipelines in a network, and some pipelines must be full. The network contains the source vertex labeled 1 and the sink vertex labeled N. Find the smallest stream in the network flow.

Practice: A general upstream/downstream network stream can be used to test the template.

# Include <stdio. h> # include <string. h> # define LMT 200 # define EPS 2e9typedef struct {int U, V;} line; line E [LMT * LMT]; int mat [LMT] [LMT], fmat [LMT] [LMT], Lim [LMT] [LMT], lev[ LMT], Q [LMT]; int n, m, sum; int BFS (INT s, int t) {int I, j, tail, head; memset (lev, 0, sizeof (lev)); lev[ S] = 1; head = tail = 0; Q [tail ++] = s; while (Head <tail) {I = Q [head ++]; for (j = 0; j <= n + 1; j ++) if (fmat [I] [J]> 0 & 0 = lev[ J]) {lev[ J] = lev[ I] + 1; Q [tail ++] = J;} Return lev[ T]! = 0;} int DFS (int s, int t) {int I, j, Top = 0, ret = 0; Q [top ++] = s; while (top> 0) {I = Q [Top-1]; if (I = T) {int min, back; min = EPS; for (I = 1; I <top; I ++) {If (min> fmat [Q [I-1] [Q [I]) {min = fmat [Q [I-1] [Q [I]; back = I ;}} RET + = min; for (I = 1; I <top; I ++) {fmat [Q [I-1] [Q [I]-= min; fmat [Q [I] [Q [I-1] + = min ;} top = back;} else {for (j = 0; j <= n + 1; j ++) if (fmat [I] [J]> 0 & lev[ J] = lev[ I] + 1) {q [top ++] = J; break ;} if (j> n + 1) {[I] = 0; top -- ;}} return ret;} int dinic (INT S, int t) {int ret = 0; while (BFS (S, T )) RET + = DFS (S, T); return ret;} int test (int s, int T, int mid) {int tn; sum = 0; mat [N] [1] = mid; memset (fmat, 0, sizeof (fmat); int I, j; for (I = 1; I <= N; I ++) {tn = 0; For (j = 1; j <= N; j ++) {tn + = lim [J] [I]-lim [I] [J]; fmat [I] [J] = mat [I] [J]-lim [I] [J];} If (TN> 0) {fmat [s] [I] = tn; sum + = tn;} else fmat [I] [T] =-Tn;} return dinic (S, T)> = sum;} in T Main () {int L, R, I, j, c, d, mid, X, ans; scanf ("% d", & N, & M ); memset (MAT, 0, sizeof (MAT); memset (Lim, 0, sizeof (lim); r = 0; L = 0; // here, l must be 0; otherwise, an error may occur. Somehow, the displayed PE for (x = 0; x <m; X ++) {scanf ("% d", & I, & J, & C, & D); R + = C; mat [I] [J] = C; If (d) Lim [I] [J] = C; E [X]. U = I; E [X]. V = J;} ans =-1; while (L <= r) {mid = (L + r)> 1; if (test (0, n + 1, mid) {r = mid-1; ans = mid;} else l = Mid + 1;} If (ANS =-1) {puts ("impossible "); return 0;} e LSE {test (0, n + 1, ANS); printf ("% d \ n", ANS); for (I = 0; I <m; I ++) printf ("% d % C", mat [E [I]. u] [E [I]. v]-fmat [E [I]. u] [E [I]. v], I = S-1? '\ N': '');} return 0 ;}