UVA

题意

给定一些插头设备和插座，有一些方法可以把其中一些插头变成另一种插头。求无法匹配插座的插头设备个数。

题解

用(map)给每个字符串标号为(a_i)和(b_i)。

读入每种改变插头的方法，连边，权值为(inf)。

然后连边(S longrightarrow a_i)，权值为(1)；(b_i longrightarrow T)，权值为(1)。

跑最大流即可。

代码

#include <bits/stdc++.h>

#define FOPI freopen("in.txt", "r", stdin)
#define FOPO freopen("out.txt", "w", stdout)
#define FOR(i,x,y) for (int i = x; i <= y; i++)
#define ROF(i,x,y) for (int i = x; i >= y; i--)

using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 100;
const int maxm = 2e5 + 100;


struct Edge
{
    int to, next, cap, flow;
}edge[maxm];

int tot;
int head[maxn];

void init()
{
    tot = 2;
    memset(head, -1, sizeof(head));
}

void build(int u, int v, int w, int rw = 0)
{
    edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0;
    edge[tot].next = head[u]; head[u] = tot++;

    edge[tot].to = u; edge[tot].cap = 0; edge[tot].flow = 0;
    edge[tot].next = head[v]; head[v] = tot++;
}

int Q[maxn];
int dep[maxn], cur[maxn], sta[maxn];

bool bfs(int s, int t, int n)
{
    int front = 0, tail = 0;
    memset(dep, -1, sizeof(dep[0]) * (n+1));
    dep[s] = 0;
    Q[tail++] = s;
    while(front < tail)
    {
        int u = Q[front++];
        for (int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if (edge[i].cap > edge[i].flow && dep[v] == -1)
            {
                dep[v] = dep[u] + 1;
                if (v == t) return true;
                Q[tail++] = v;
            }
        }
    }
    return false;
}

LL dinic(int s, int t, int n)
{
    LL maxflow = 0;
    while(bfs(s, t, n))
    {
        for (int i = 0; i < n; i++) cur[i] = head[i];
        int u = s, tail = 0;
        while(cur[s] != -1)
        {
            if (u == t)
            {
                int tp = inf;
                for (int i = tail-1; i >= 0; i--)
                    tp = min(tp, edge[sta[i]].cap - edge[sta[i]].flow);

                //if (tp >= inf) return -1;
                maxflow += tp;

                for (int i = tail-1; i >= 0; i--)
                {
                    edge[sta[i]].flow += tp;
                    edge[sta[i]^1].flow -= tp;
                    if (edge[sta[i]].cap - edge[sta[i]].flow == 0) tail = i;
                }
                u = edge[sta[tail]^1].to;
            }
            else if (cur[u] != -1 && edge[cur[u]].cap > edge[cur[u]].flow
                     && dep[u]+1 == dep[edge[cur[u]].to])
            {
                sta[tail++] = cur[u];
                u = edge[cur[u]].to;
            }
            else
            {
                while(u != s && cur[u] == -1) u = edge[sta[--tail]^1].to;
                cur[u] = edge[cur[u]].next;
            }
        }
    }
    return maxflow;
}

int t;
int a[205], b[205];
string s, r;
map<string, int> M;
int n, m, S, T, cnt, q;

int getid(string &s)
{
    if (M.count(s)) return M[s];
    return M[s] = ++cnt;
}

int main()
{
//    FOPI;
//    FOPO;

    ios::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> t;
    FOR(ca, 1, t)
    {
        init();
        M.clear(); cnt = 0;

        cin >> n;
        FOR(i, 1, n) { cin >> s; a[i] = getid(s); }

        cin >> m;
        FOR(i, 1, m) { cin >> r >> s; b[i] = getid(s); }

        cin >> q;
        FOR(i, 1, q)
        {
            cin >> s >> r;
            int x = getid(s), y = getid(r);
            build(y, x, inf);
        }

        S = 0, T = cnt+1;
        FOR(i, 1, n) build(S, a[i], 1);
        FOR(j, 1, m) build(b[j], T, 1);

        LL ans = dinic(S, T, T);
        printf("%lld
", m-ans);

        if (ca != t) puts("");
    }
}