[USACO09JAN]Total Flow S【最大流模板題】

題目鏈接 P2936 [USACO09JAN]Total Flow S

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約翰總希望他的奶牛有足夠的水喝，因此他找來了農場的水管地圖，想算算牛棚得到的水的 總流量.農場裏一共有N根水管.約翰發現水管網絡混亂不堪，他試圖對其進行簡 化.他簡化的方式是這樣的：

兩根水管串聯，則可以用較小流量的那根水管代替總流量.

兩根水管並聯，則可以用流量爲兩根水管流量和的一根水管代替它們

當然，如果存在一根水管一端什麼也沒有連接，可以將它移除.

請寫個程序算出從水井A到牛棚Z的總流量.數據保證所有輸入的水管網絡都可以用上述方法 簡化.

  那麼，這道題就是一道網絡流最大流的模板題了，直接利用源點爲A點，匯點爲Z點，然後跑一個最大流即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define eps 1e-8
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
#define MAX_3(a, b, c) max(a, max(b, c))
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e2, maxM = 2e3 + 7;
int N, tot = 0;
int S, T, head[maxN], cnt, cur[maxN];
unordered_map<char, int> mp;
struct Eddge
{
    int nex, to; ll flow;
    Eddge(int a=-1, int b=0, ll c=0):nex(a), to(b), flow(c) {}
}edge[maxM];
inline void addEddge(int u, int v, ll w)
{
    edge[cnt] = Eddge(head[u], v, w);
    head[u] = cnt++;
}
inline void _add(int u, int v, ll w) { addEddge(u, v, w); addEddge(v, u, 0); }
struct Max_Flow
{
    int gap[maxN], d[maxN], que[maxN], ql, qr, node;
    inline void init()
    {
        for(int i=0; i<=node + 1; i++)
        {
            gap[i] = d[i] = 0;
            cur[i] = head[i];
        }
        ++gap[d[T] = 1];
        que[ql = qr = 1] = T;
        while(ql <= qr)
        {
            int x = que[ql ++];
            for(int i=head[x], v; ~i; i=edge[i].nex)
            {
                v = edge[i].to;
                if(!d[v]) { ++gap[d[v] = d[x] + 1]; que[++qr] = v; }
            }
        }
    }
    inline ll aug(int x, ll FLOW)
    {
        if(x == T) return FLOW;
        int flow = 0;
        for(int &i=cur[x], v; ~i; i=edge[i].nex)
        {
            v = edge[i].to;
            if(d[x] == d[v] + 1)
            {
                ll tmp = aug(v, min(FLOW, edge[i].flow));
                flow += tmp; FLOW -= tmp; edge[i].flow -= tmp; edge[i ^ 1].flow += tmp;
                if(!FLOW) return flow;
            }
        }
        if(!(--gap[d[x]])) d[S] = node + 1;
        ++gap[++d[x]]; cur[x] = head[x];
        return flow;
    }
    inline ll max_flow()
    {
        init();
        ll ret = aug(S, INF);
        while(d[S] <= node) ret += aug(S, INF);
        return ret;
    }
} mf;
inline void init()
{
    cnt = 0; S = 1; T = 2; mf.node = 52;
    for(int i=0; i<=mf.node; i++) head[i] = -1;
    mp['A'] = ++tot; mp['Z'] = ++tot;
}
int main()
{
    scanf("%d", &N);
    init();
    for(int i=1; i<=N; i++)
    {
        char u[3], v[3]; int fi;
        scanf("%s%s%d", u, v, &fi);
        if(!mp[u[0]]) mp[u[0]] = ++tot;
        if(!mp[v[0]]) mp[v[0]] = ++tot;
        _add(mp[u[0]], mp[v[0]], fi);
    }
    printf("%lld\n", mf.max_flow());
    return 0;
}