POJ 1087 A Plug for UNIX (最大流，EK演算法）

題意：讀題很噁心···。大概就是說現在有n個不同的插孔，m台不同的用電器，k種適配器。（適配器就相當於一個中間插座，比如一個適配器是(x,y)。有一台用電器必須插x插孔，但是現在只有一個y插孔，那麼就可以通過適配器來串連）。另外需要注意的就是給出的是適配器的種數，每一種的數量無限制。
題解：建圖的大概思路。s-->電器-->適配器-->插孔-->t

#include <queue>#include <cstring>#include <iostream>using namespace std;#define N 101#define INF 9999char recep[N][25];char device[N][25];struct Item{char in[25];char out[25];} adapt[N];bool vis[N*3];int pre[N*3];int cap[N*3][N*3];int flow[N*3][N*3];int n, m, k;int s, t, res;void build_map(){s = 0;t = n + m + k + 1;memset(cap,0,sizeof(cap));int i, j;for ( i = 1; i <= m; i++ ){cap[s][i] = 1;  // s到電器for ( j = 1; j <= n; j++ ) //電器到插孔if ( ! strcmp(device[i],recep[j]) )cap[i][j+m+k] = 1;for ( j = 1; j <= k; j++ )  //電器到適配器if ( ! strcmp(device[i],adapt[j].in) )cap[i][j+m] = 1;}for ( i = 1; i <= k; i++ ){for ( j = 1; j <= k; j++ )  // 適配器之間聯絡    if ( i != j && !strcmp(adapt[i].out, adapt[j].in) )cap[i+m][j+m] = INF;   // 數量不限制，所以是INFfor ( j = 1; j <= n; j++ )  // 適配器到插孔    if ( ! strcmp(adapt[i].out, recep[j]) )cap[i+m][j+m+k] = 1;}for ( i = 1; i <= n; i++ )  //插孔到tcap[i+m+k][t] = 1;}bool find_path () // BFS找增廣路{memset(pre,-1,sizeof(-1));memset(vis,0,sizeof(vis));queue<int> que;vis[s] = 1;que.push(s);while ( ! que.empty () ){int u = que.front ();que.pop ();for ( int v = t; v >= s; v-- ){if ( ! vis[v] && cap[u][v] > flow[u][v] ){    vis[v] = 1;    pre[v] = u;    if ( v == t ) return true;    que.push(v);}}}return false;}int max_flow(){res = 0;memset(flow,0,sizeof(flow));while ( 1 ){if ( ! find_path() ) break;int tt = t;int minFlow = INF;while ( pre[tt] != -1 ){if ( minFlow > cap[pre[tt]][tt] - flow[pre[tt]][tt] )minFlow = cap[pre[tt]][tt] - flow[pre[tt]][tt];tt = pre[tt];}tt = t;while ( pre[tt] != -1 ){flow[pre[tt]][tt] += minFlow;flow[tt][pre[tt]] -= minFlow;tt = pre[tt];}res += minFlow;}return res;}int main(){int i; char name[25];scanf("%d",&n); for ( i = 1; i <= n; i++ )scanf("%s",recep[i]);scanf("%d",&m); for ( i = 1; i <= m; i++ )scanf("%s %s", name, device[i]);scanf("%d",&k);for ( i = 1; i <= k; i++ )scanf("%s %s",adapt[i].in, adapt[i].out);build_map();printf("%d\n", m - max_flow());return 0;}