HDU 4940 Destroy Transportation System (feasible flow judgment without Yuanhui upper and lower bounds network flow)

Test instructions: There are n points and m bars have a network of edges, each side has two costs:

D: Cost of destroying this side B: rebuilding the cost of a two-way side

Looking for such two point sets, so that the point set S points set T satisfies the cost of destroying all the paths of S to T and > destroys all T to S of the cost of the path and + rebuilds these T to s the cost of the bidirectional path and.

Idea 1:

And then this is the solution to the feasible flow problem of the Yuanhui with the Nether network flow see here ~ ~

This is what it says. Finally, there is no viable flow is to find the maximum flow of the new network we construct ~ and then determine whether the maximum flow is full flow ~ (that is, determine whether the maximum flow and the additional source points of the total outflow equal ~ ~)

Code

#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include < vector> #include <string> #include <queue> #include <map> #include <set> #include <cmath > #include <cstdlib>using namespace std, #define INF 0x3f3f3f3f#define PI acos ( -1.0) #define MEM (A, B) memset (A, B, sizeof (a)) typedef pair<int,int> PII;TYPEDEF Long Long ll;//------------------------------const int MAXN = 210;int n,m;struct edge{int to, CAP, rev;//end capacity reverse edge edge (int to_ = 0, int cap_ = 0, int rev_ = 0) {to = To_            ;            Cap = CAP_;      Rev = rev_; }};vector<edge> g[maxn];int level[maxn];//Vertex to source point distance designator int iter[maxn];void Add_edge (int from, int. to, int cap) {Ed      GE Tmp1 (To, Cap, g[to].size ());      G[from].push_back (TMP1);      Edge tmp2 (from, 0, G[from].size ()-1); G[to].push_back (TMP2);}      void BFs (int s) {memset (level,-1, sizeof);      Queue<int> Q;      Level[s] = 0; Q.push (s);      while (!q.empty ()) {int v = q.front ();            Q.pop ();                  for (int i = 0; i < g[v].size (); i++) {edge& e = g[v][i];                  if (E.cap > 0 && level[e.to] < 0) {level[e.to] = Level[v] + 1;                  Q.push (e.to);      }}}}int dfs (int v, int t, int f) {if (v = = t) return F;            for (int& i = iter[v]; i < g[v].size (); i++) {edge& e = g[v][i];                  if (E.cap > 0 && level[v] < level[e.to]) {int d = DFS (e.to, T, Min (E.cap, f));                  if (d > 0) {e.cap-= D;                  G[e.to][e.rev].cap + = D;                  return D; }}} return 0;}      int Max_flow (int s, int t) {int flow = 0; for (;;)            {BFS (s);            if (Level[t] < 0) return flow;            memset (ITER, 0, sizeof (ITER));            int F; while ((f = DFS (s, T, INF)) > 0) {flow + = f; }}}//------------------above is the maximum flow template part correct----------------------int Sum;int mi[maxn];//This is the value of the degrees-out that is used to hold each point, which is the M (i) in the paper; VO      ID init () {for (int i = 0; i < MAXN; i++) {//Net stream empty, do not forget g[i].clear ();      } memset (MI, 0, sizeof (MI));      scanf ("%d%d", &n,&m);      int U, V, d, b;            for (int i = 1; I <= m; i++) {scanf ("%d%d%d%d", &u, &v, &d, &b); Mi[u]-=d;            Mi[v]+=d; Add_edge (U, V, b);//plus Edge added is the nether value of each edge} sum = 0;//sum All outgoing traffic for the source point for (int i = 1; I <= n; i++) {i            F (Mi[i] < 0) Add_edge (i, n+1,-mi[i]);                  else if (Mi[i] > 0) {add_edge (0, I, mi[i]);            Sum + = Mi[i];      }}}void Solve () {int ans = max_flow (0, n+1);      if (ans = = sum) printf ("happy\n"); else printf ("unhappy\n");}      int main () {int t;      int cas = 0;      scanf ("%d", &t); while (t--){init ();            printf ("Case #%d:", ++cas);      Solve (); } return 0;}

Idea 2:

To imagine ~ If for all a single point of a set of S, can not meet

The cost of the path of S to T and > destroys all T to S of the cost of the path and the cost of rebuilding these T to S bidirectional path, then the collection of arbitrary points constitutes the S set also must not meet the above conditions ~ ~ ~ If the existence of a set satisfies the above conditions, there must be a single point to meet the above conditions ~~~~

So we actually just use the enumeration point and then judge it. Orz ...

Code

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include < vector>using namespace Std;const int maxn = 205;int N, m;int D[maxn][maxn], b[maxn][maxn];vector<int> IN[MAXN], O            Ut[maxn];void Init () {for (int i = 0; i < MAXN; i++) {in[i].clear ();      Out[i].clear ();      } memset (D,0,sizeof (D));      memset (b,0,sizeof (B));      scanf ("%d%d", &n,&m);      int u, V, DD, BB;            for (int i = 1; I <= m; i++) {scanf ("%d%d%d%d", &AMP;U,&AMP;V,&AMP;DD,&AMP;BB);            Out[u].push_back (v);            In[v].push_back (U);            D[U][V] = DD;      B[U][V] = BB;      }}bool deal (int u) {int x = 0;            for (int i = 0; i < in[u].size (); i++) {int v = in[u][i];      x + = D[v][u];      } int y = 0;            for (int i = 0; i < out[u].size (); i++) {int v = out[u][i];      y = y + d[u][v] + b[u][v];      } if (Y < x) return true; RetuRN false;}                  void Solve () {for (int i = 1; I <= n; i++) {if (Deal (i)) {printf ("unhappy\n");            Return }} printf ("happy\n");}      int main () {int t;      int cas = 0;      scanf ("%d", &t);            while (t--) {init ();            printf ("Case #%d:", ++cas);      Solve (); } return 0;}

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HDU 4940 Destroy Transportation System (feasible flow judgment without Yuanhui upper and lower bounds network flow)