POJ 1087 A Plug for UNIX

链接：http://poj.org/problem?id=1087
题意：有n（1≤n≤100）个插座，每个插座用一个数字字母式字符串描述（至多有24 个字符）。有m（1≤m≤100）个设备，每个设备有名称，以及它使用的插头的名称；插头的名称跟它所使用的插座的名称是一样的；设备名称是一个至多包含24 个字母数字式字符的字符串；任何两个设备的名称都不同；有k（1≤k≤100）个转换器，每个转换器能将插座转换成插头。
样例：

4 
A 
B 
C 
D 
5 
laptop B 
phone C 
pager B 
clock B 
comb X 
3 
B X 
X A 
X D

思路：建立一个汇点，每个插座和汇点连边，流量代表该种插座的个数。建立一个源点，源点和每个设备连边，流量为1。每个转换器将两个插座连边，流量为INF。然后求最大流。答案就是设备的个数减去最大流。
细节：这道题在建图的时候比较恶心。需要注意的是，在m个设备和插座以及k个转换器的数据中，可能有之前没有出现过的设备。所以必须将插座的信息存起来，可以用map来存插座对应的点。
代码：

/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include
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                using namespace std; #define INF (1<<30) #define PI acos(-1.0) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i < n; i++) #define per(i, a, n) for (int i = n - 1; i >= a; i--) #define eps 1e-6 #define debug puts("===============") #define pb push_back //#define mp make_pair #define all(x) (x).begin(),(x).end() #define fi first #define se second #define SZ(x) ((int)(x).size()) #define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m) typedef long long ll; typedef unsigned long long ULL; const int maxn = 610; const int maxm = 2000000; int m, k; struct node { int v; // vertex int cap; // capacity int flow; // current flow in this arc int nxt; } e[maxm * 2]; int g[maxn], cnt; int st, ed, n; void add(int u, int v, int c) { e[++cnt].v = v; e[cnt].cap = c; e[cnt].flow = 0; e[cnt].nxt = g[u]; g[u] = cnt; e[++cnt].v = u; e[cnt].cap = 0; e[cnt].flow = 0; e[cnt].nxt = g[v]; g[v] = cnt; } map
               
                 mp; map
                
                  :: iterator it; void init() { mem(g, 0); cnt = 1; mp.clear(); scanf("%d", &n); char str[25]; ed = 0; int tot = 1, has[111] = {0}; for (int i = 1; i <= n; i++) { scanf("%s", str); string ss(str); int now; if (mp[ss] == 0) { mp[ss] = tot; now = tot++; } else now = mp[ss]; has[now]++; } for (int i = 1; i < tot; i++) add(i, ed, has[i]); scanf("%d", &m); char s[25]; vector
                 
                   v; for (int i = 0; i < m; i++) { scanf("%s%s", str, s); v.pb(tot); if (mp[s]) add(tot++, mp[s], 1); else { mp[s] = tot + 1; add(tot, tot + 1, 1); tot += 2; } } scanf("%d", &k); for (int i = 0; i < k; i++) { scanf("%s%s", str, s); if (mp[str] == 0) mp[str] = tot++; if (mp[s] == 0) mp[s] = tot++; add(mp[str], mp[s], INF); } st = tot; for (int i = 0; i < v.size(); i++) add(st, v[i], 1); n = tot + 3; } int dist[maxn], numbs[maxn], q[maxn]; void rev_bfs() { int font = 0, rear = 1; for (int i = 0; i <= n; i++) { //n为总点数 dist[i] = maxn; numbs[i] = 0; } q[font] = ed; dist[ed] = 0; numbs[0] = 1; while(font != rear) { int u = q[font++]; for (int i = g[u]; i; i = e[i].nxt) { if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue; dist[e[i].v] = dist[u] + 1; ++numbs[dist[e[i].v]]; q[rear++] = e[i].v; } } } int maxflow() { rev_bfs(); int u, totalflow = 0; int curg[maxn], revpath[maxn]; for(int i = 0; i <= n; ++i) curg[i] = g[i]; u = st; while(dist[st] < n) { if(u == ed) { // find an augmenting path int augflow = INF; for(int i = st; i != ed; i = e[curg[i]].v) augflow = min(augflow, e[curg[i]].cap); for(int i = st; i != ed; i = e[curg[i]].v) { e[curg[i]].cap -= augflow; e[curg[i] ^ 1].cap += augflow; e[curg[i]].flow += augflow; e[curg[i] ^ 1].flow -= augflow; } totalflow += augflow; u = st; } int i; for(i = curg[u]; i; i = e[i].nxt) if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break; if(i) { // find an admissible arc, then Advance curg[u] = i; revpath[e[i].v] = i ^ 1; u = e[i].v; } else { // no admissible arc, then relabel this vertex if(0 == (--numbs[dist[u]])) break; // GAP cut, Important! curg[u] = g[u]; int mindist = n; for(int j = g[u]; j; j = e[j].nxt) if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]); dist[u] = mindist + 1; ++numbs[dist[u]]; if(u != st) u = e[revpath[u]].v; // Backtrack } } return totalflow; } int main () { init(); printf("%d\n", m - maxflow()); return 0; }


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