ZOJ1157, POJ1087,UVA 753 A Plug for UNIX (最大流)

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

题目意思有点儿难描述 用一个别人描述好的。

我的建图方法：一个源点一个汇点，和所有种类的插座。输入的n个插座直接与源点相连，容量为1，m个物品输入里 记录每个插座对应的物品个数，物品数然后大于0的插座直接连到汇点，意味着最终的物品只能由这些插座流出。中间的插座转换容量都是INF  a b表示  无论多少b都可以选择转化到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) int (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) int (X),(Y),(Z);scanf("%d%d%d",&(X),&(Y),&(Z))#define RIII(X,Y,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 < 300; 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;}