Poj 3469 dual core CPU (minimum cut)

Poj 3469 dual core CPU Link: http://poj.org/problem? Id = 3469
Question: There are two CPUs and N modules. Each module runs on a single CPU. There is m for information interaction between modules. If the module is on the same CPU, no additional time is required for information interaction; otherwise, additional time is required. Ask how to allocate modules to minimize the time spent.
Ideas: To divide the module into two CPUs, and at the same time to minimize the time, it is to find the minimum cut. Then the question can be converted to the maximum stream.
Code:

/* ID: [email protected] prog: Lang: c ++ */# include <map> # include <set> # include <queue> # include <stack> # include <cmath> # include <cstdio> # include <vector> # include <string> # include <fstream> # include <cstring> # include <ctype. h> # include <iostream> # include <algorithm> using namespace STD; # define linf (1ll <60) # define Inf (1 <30) # define PI ACOs (-1.0) # define MEM (a, B) memset (a, B, sizeof (A) # define rep (I, A, n) for (INT I = A; I <n; I ++) # define per (I, A, n) for (INT I = n-1; I> = A; I --) # define EPS 1e-6 # define debug puts ("================ ") # define Pb push_back // # define MP make_pair # define all (x ). begin (), (x ). end () # define Fi first # define se second # define SZ (x) (INT) (x ). size () # define posin (x, y) (0 <= (x) & (x) <n & 0 <= (y) & (y) <m) typedef long ll; typedef unsigned long ull; const int maxn = 1210 00; const int maxm = 1410000; int St, Ed, n, m; struct node {int V; // vertex int cap; // capacity int flow; // current flow in this arc int NXT;} e [maxm * 2]; int G [maxn], CNT; void add (int u, int V, int C) {e [++ CNT]. V = V; E [CNT]. CAP = C; E [CNT]. flow = 0; E [CNT]. NXT = G [u]; G [u] = CNT; E [++ CNT]. V = u; E [CNT]. CAP = 0; E [CNT]. flow = 0; E [CNT]. NXT = G [v]; G [v] = CNT;} void Init () {MEM (G, 0); CNT = 1; St = 0; ED = n + 1; int U, V, C; For (INT I = 1; I <= N; I ++) {scanf ("% d", & U, & V); add (St, I, U); add (I, Ed, V );} while (M --) {scanf ("% d", & U, & V, & C); add (u, v, c); add (V, u, c) ;}n = N + 3 ;}int Dist [maxn], numbs [maxn], Q [maxn]; void rev_bfs () {int font = 0, rear = 1; for (INT I = 0; I <= N; I ++) {// n indicates the total number of DIST [I] = maxn; numbs [I] = 0;} Q [font] = Ed; Dist [ed] = 0; numbs [0] = 1; while (F ONT! = Rear) {int u = Q [font ++]; for (INT I = G [u]; I; I = E [I]. NXT) {If (E [I ^ 1]. CAP = 0 | Dist [E [I]. v] <maxn) continue; Dist [E [I]. v] = DIST [u] + 1; ++ numbs [Dist [E [I]. v]; Q [rear ++] = E [I]. V ;}}int maxflow () {rev_bfs (); int U, totalflow = 0; int curg [maxn], revpath [maxn]; for (INT I = 0; I <= N; ++ I) curg [I] = G [I]; u = sT; while (Dist [st] <n) {If (u = ed) {// find an Augmenting Path int augf Low = inf; For (INT I = sT; I! = Ed; I = E [curg [I]. v) augflow = min (augflow, E [curg [I]. CAP); For (INT I = sT; I! = Ed; I = E [curg [I]. v) {e [curg [I]. cap-= augflow; E [curg [I] ^ 1]. cap + = augflow; E [curg [I]. flow + = augflow; E [curg [I] ^ 1]. flow-= augflow;} totalflow + = augflow; u = sT;} int I; for (I = curg [u]; I; I = E [I]. NXT) if (E [I]. cap> 0 & Dist [u] = DIST [E [I]. v] + 1) break; if (I) {// find an admissible arc, then advance curg [u] = I; revpath [E [I]. v] = I ^ 1; u = E [I]. v;} else {// no admissible arc, Then relabel This vertex if (0 = (-- numbs [Dist [u]) break; // gap cut, important! Curg [u] = G [u]; int mindist = N; For (Int J = G [u]; j = E [J]. NXT) if (E [J]. cap> 0) mindist = min (mindist, DIST [E [J]. v]); Dist [u] = mindist + 1; ++ numbs [Dist [u]; If (u! = ST) u = E [revpath [u]. V; // backtrack} return totalflow;} int main () {While (~ Scanf ("% d", & N, & M) {Init (); printf ("% d \ n", maxflow () ;}return 0 ;}

Poj 3469 dual core CPU (minimum cut)