POJ--1087（最大流）
2014-12-20 15:58:19
思路：考了一发建图....其实也比较简单，首先定一个总源点和一个总汇点，然后根据拥有的插座建立中间点到汇点的通道，容量为1，根据需求建立源点到中间点的通道，容量为某种插座的需求量。对于转换器，在中间点之间建立容量为INF的通道即可。
（others：当然这题这种建图可以完全倒过来，相当于从总汇点倒流到总源点，那么就要把总汇/源点互换，转换器的建的边反一下。）
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cmath> 5 #include <vector> 6 #include <map> 7 #include <set> 8 #include <stack> 9 #include <queue> 10 #include <iostream> 11 #include <algorithm> 12 using namespace std; 13 #define lp (p << 1) 14 #define rp (p << 1|1) 15 #define getmid(l,r) (l + (r - l) / 2) 16 #define MP(a,b) make_pair(a,b) 17 typedef long long ll; 18 typedef unsigned long long ull; 19 const int INF = 1 << 30; 20 const int maxn = 410; 21 22 int n,m,k,tot; 23 string s[maxn]; 24 map<string,int> mp; 25 int c[maxn][maxn],lev[maxn]; 26 int Q[maxn],head,rear; 27 28 void Bfs(){ 29 memset(lev,-1,sizeof(lev)); 30 lev[0] = 0; 31 Q[head = rear = 1] = 0; 32 while(head <= rear){ 33 int x = Q[head++]; 34 for(int i = 1; i <= tot; ++i) if(c[x][i] > 0 && lev[i] < 0){ 35 lev[i] = lev[x] + 1; 36 Q[++rear] = i; 37 } 38 } 39 } 40 41 int Dfs(int p,int minf){ 42 if(p == tot) return minf; 43 for(int i = 1; i <= tot; ++i) if(lev[i] > lev[p] && c[p][i] > 0){ 44 int d = Dfs(i,min(minf,c[p][i])); 45 if(d > 0){ 46 c[p][i] -= d; 47 c[i][p] += d; 48 return d; 49 } 50 } 51 return 0; 52 } 53 54 int Dinic(){ 55 int max_flow = 0,plus; 56 while(1){ 57 Bfs(); 58 if(lev[tot] < 0) break; 59 while((plus = Dfs(0,INF)) > 0) max_flow += plus; 60 } 61 return max_flow; 62 } 63 64 int main(){ 65 memset(c,0,sizeof(c)); 66 tot = 0; 67 string t1,t2; 68 scanf("%d",&n); 69 for(int i = 1; i <= n; ++i){ 70 cin >> s[i]; 71 mp[s[i]] = ++tot; 72 } 73 scanf("%d",&m); 74 for(int i = 1; i <= m; ++i){ 75 cin >> t1 >> t2; 76 if(mp.find(t2) == mp.end()) mp[t2] = ++tot; 77 c[0][mp[t2]]++; 78 } 79 scanf("%d",&k); 80 for(int i = 1; i <= k; ++i){ 81 cin >> t1 >> t2; 82 if(mp.find(t1) == mp.end()) mp[t1] = ++tot; 83 if(mp.find(t2) == mp.end()) mp[t2] = ++tot; 84 c[mp[t1]][mp[t2]] = INF; 85 } 86 ++tot; 87 for(int i = 1; i <= n; ++i) 88 c[mp[s[i]]][tot] = 1; 89 printf("%d\n",m - Dinic()); 90 return 0; 91 }
