POJ 1087 A Plug for UNIX （网络最大流）

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: [e-mail]:LANG: C++*/#include<map>#include<set>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<fstream>#include<cstring>#include<ctype.h>#include<iostream>#include<algorithm>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<string, int> mp;map<string, int> :: 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<int> 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;}