ZOJ1157, Poj1087,uva 753 A Plug for UNIX (Max Stream)

Links: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=26746

The topic meaning is a bit difficult to describe with a person described well.

My map method: One source point, one sink point, and all kinds of sockets. The n sockets are connected directly to the source point, the capacity is the number of items per outlet in the 1,m item input, and the number of items greater than 0 is connected directly to the meeting point, which means that the final item can only be drained from these outlets. The middle of the socket conversion capacity is INF a B means that no matter how many B can choose to convert to a.

/*---------------------#headfile--------------------*/#include <algorithm> #include <iostream># Include <sstream> #include <cstring> #include <cstdlib> #include <cassert> #include <cstdio > #include <vector> #include <cmath> #include <queue> #include <stack> #include <set># Include <map>/*----------------------#define----------------------*/#define DRII (x, y), scanf ("%d% D ",& (X),& (Y)) #define EXP 2.7182818284590452353602874713527#define caset int _;cin>>_;while (_--) # Define RII (x, Y) scanf ("%d%d",& (x),& (Y)) #define DRI (x) int (x), scanf ("%d", &x) #define MEM (A, B) memset (A, B, sizeof (a)) #define REP (I,n) for (int. i=0;i<n;i++) #define ALL (x) (x). Begin (), (x). End () #define INFL 0x3f3f3f3f3f3f3f3fll#define RI (x) scanf ("%d",& (x)) #define SZ (x) ((int) x.size ()) #define PDI Pair<double,int > #define Rson o<<1|1,m+1,r#define PII pair<int,int> #define MAX 0x3f3f3f3f#define Lson o<<1, L,m#define MP make_pair#define PB push_back#define SE second#define FI firsttypedef Long Long ll;template<class T>T MUL (t x,t y,t P) {t f1=0;while (y) {if (y&1) {f1+=x;if (f1<0| | f1>=p) f1-=p;} X<<=1;if (x<0| | x>=p) X-=p;y>>=1;} return F1;} Template<class t>t POW (t x,t y,t P) {t f1=1;x%=p;while (y) {if (y&1) F1=mul (f1,x,p); X=mul (x,x,p); y>>=1;} return F1;} Template<class t>t gcd (T x,t y) {if (y==0) return x; T z;while (z=x%y) X=y,y=z;return y;} #define DRIII (x, y), (z), scanf ("%d%d%d",& (x),& (y),& (z)) #define RIII, X, Z, scanf ("%d%d%d", & (X),& (Y),& (Z)) const double PI = ACOs ( -1.0); const double EPS = 1e-6;const ll mod = 1000000007ll;const int M =  1005;const int N = 605;using namespace std;/*----------------------Main-------------------------*/struct Edge {int to,    C, Rev;    Edge () {} edge (int _to, int _c, int _rev) {to = _to, C = _c, rev = _rev; }};vector<edge> G[n];int Lv[n], iter[n];int N, M;void BFS (inT s) {mem (LV,-1);    Queue<int> Q;    Lv[s] = 0;    Q.push (s);        while (!q.empty ()) {int v = q.front (); Q.pop ();            for (int i = 0; i < SZ (G[v]); i++) {Edge &e = g[v][i];                if (e.c > 0 && lv[e.to] < 0) {lv[e.to] = Lv[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 < SZ (G[v]); i++) {Edge &e = g[v][i];            if (e.c > 0 && lv[v] < lv[e.to]) {int d = DFS (e.to, T, Min (f, e.c));                if (d > 0) {e.c-= D;                G[E.TO][E.REV].C + = D;            return D; }}} return 0;}    int MF (int s, int t) {int res = 0; for (;;)        {BFS (s);        if (Lv[t] < 0) return res;        MEM (ITER, 0);        int F;        while ((f = DFS (S, T, 1e9)) > 0) {res + = f; }}}void Add (int from, int. to,int c) {G[from].    PB (Edge (To, C, SZ (g[to))); G[to]. PB (Edge (from, 0, SZ (G[from])-1));} int Num[n];int FF = 0;void Solve () {if (FF) puts ("");    FF = 1;    RI (n);    for (int i = 0; i <; i++) g[i].clear ();    MEM (num, 0);    int s = 0, k = 0;    Map<string, int> vis;        for (int i = 1; I <= n; i++) {string S1; Cin >> S1;        VIS[S1] = ++k;    Add (S, I, 1);    } RI (M);        for (int i = 1; I <= m; i++) {string S1, S2;        CIN >> S1 >> S2;        if (vis[s2] = = 0) vis[s2] = ++k;    num[Vis[s2]]++;    } int t = k + 1;    for (int i = 1; I <= K; i++) {if (Num[i]) Add (i, T, Num[i]);    } DRI (x);    k++;        for (int i = 1; i <= x; i++) {string S1, S2;        CIN >> S1 >> S2;        if (vis[s2] = = 0) vis[s2] = ++k;        if (vis[s1] = = 0) vis[s1] = ++k;        int u = vis[s2], v = vis[s1];    Add (U, V, 1e9);    } int ans = MF (s, t); printf ("%d\n", M-ans);}  int main () {//  Freopen ("In.txt", "R", stdin);//Freopen ("OUT.txt", "w", stdout);    Caset solve (); return 0;}

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ZOJ1157, Poj1087,uva 753 A Plug for UNIX (Max Stream)