题意:给定 n 种插座,m种设备,和k个转换器,问你最少有几台设备不能匹配。
析:一个很裸的网络流,直接上模板就行,建立一个源点s和汇点t,源点和每个设备连一条边,每个插座和汇点连一条边,然后再连转换器,
最后跑一次最大流即可。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-5; const int maxn = 500 + 10; const int mod = 1e6; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } struct Edge{ int from, to, cap, flow; }; struct Dinic{ int n, m, s, t; vector<Edge> edges; vector<int> G[maxn]; bool vis[maxn]; int d[maxn]; int cur[maxn]; void init(){ edges.clear(); for(int i = 0; i < maxn; ++i) G[i].clear(); } bool bfs(){ memset(vis, 0, sizeof vis); queue<int> q; q.push(s); d[s] = 0; vis[s] = true; while(!q.empty()){ int x = q.front(); q.pop(); for(int i = 0; i < G[x].size(); ++i){ Edge &e = edges[G[x][i]]; if(!vis[e.to] && e.cap > e.flow){ vis[e.to] = 1; d[e.to] = d[x] + 1; q.push(e.to); } } } return vis[t]; } int dfs(int x, int a){ if(x == t || a == 0) return a; int flow = 0, f; for(int& i = cur[x]; i < G[x].size(); ++i){ Edge& e = edges[G[x][i]]; if(d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){ e.flow += f; edges[G[x][i]^1].flow -= f; flow += f; a -= f; if(!a) break; } } return flow; } int maxflow(int s, int t){ this->s = s; this->t = t; int flow = 0; while(bfs()){ memset(cur, 0, sizeof cur); flow += dfs(s, INF); } return flow; } void addEdge(int from, int to, int cap){ edges.push_back(Edge{from, to, cap, 0}); edges.push_back(Edge{to, from, 0, 0}); m = edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } }; map<string, int> mp; int getId(const string &s){ if(mp.count(s)) return mp[s]; return mp[s] = mp.size(); } vector<int> socket; vector<P> device; vector<P> convert; void init(){ mp.clear(); socket.clear(); device.clear(); convert.clear(); } int solve(){ int s = mp.size() + 1; int t = mp.size() + 2; Dinic dinic; dinic.init(); for(int i = 0; i < device.size(); ++i){ dinic.addEdge(s, device[i].first, 1); dinic.addEdge(device[i].first, device[i].second, 1); } for(int i = 0; i < socket.size(); ++i) dinic.addEdge(socket[i], t, 1); for(int i = 0; i < convert.size(); ++i) dinic.addEdge(convert[i].first, convert[i].second, INF); return (int)device.size() -dinic.maxflow(s, t); } int main(){ int T; cin >> T; while(T--){ cin >> n; init(); string s; for(int i = 0; i < n; ++i){ cin >> s; socket.push_back(getId(s)); } cin >> n; string ss; for(int i = 0; i < n; ++i){ cin >> s >> ss; device.push_back(P(getId(s), getId(ss))); } cin >> n; for(int i = 0; i < n; ++i){ cin >> s >> ss; convert.push_back(P(getId(s), getId(ss))); } cout << solve() << endl; if(T) cout << endl; } return 0; }