Flow construction 【SGU - 176】【有源汇有上下界最小流+负流特殊考虑】

Flow construction

 SGU - 176

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  有N个点，M条边，每条边有{u, v, w, c}，如果c==1表示这条边一定要满流流过去，否则，这条边的流量可以是0～w的任意。求最小可行流。

  于是，我们很简单的构造了一个可行流网络，这些都是简单的，关键是求出可行流之后的步骤，有很多的细节上的东西。

  首先，题目中给的源汇点是1和N，但是如果有直接从1到N的边，会形成环流这样的形式，作为DAG图的网络流显然是不允许这样的情况出现的，所以，我们开始的时候独立创建一个s、t的源汇点。s链接1号点，N号点链接t，流量为INF，无上下界。然后，我们就避免了源汇点之间构成了直接环的情况，这时候再去跑就满足网络流的条件了。

  现在，既然是要求最小流，那么我们将起点设置为t，将终点设置为s，相当于反过来了，同时不要忘记删除我们加入的t到s的“使有源汇图看作无源汇图”的这条特殊边。

#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 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
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 107, maxM = 2e4 + 7;
int N, M, head[maxN], cnt;
ll du[maxN], fl[maxM], sum;
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 f)
{
    edge[cnt] = Eddge(head[u], v, f);
    head[u] = cnt++;
}
inline void _add(int u, int v, ll f) { addEddge(u, v, f); addEddge(v, u, 0); }
struct ISAP
{
    int S, T, gap[maxN], cur[maxN], deep[maxN], que[maxN], ql, qr, node;
    inline void init()
    {
        for(int i=0; i<=node + 1; i++)
        {
            gap[i] = deep[i] = 0;
            cur[i] = head[i];
        }
        ++gap[deep[T] = 1];
        que[ql = qr = 1] = T;
        while(ql <= qr)
        {
            int u = que[ql ++];
            for(int i=head[u], v; ~i; i=edge[i].nex)
            {
                v = edge[i].to;
                if(!deep[v]) { ++gap[deep[v] = deep[u] + 1]; que[++qr] = v; }
            }
        }
    }
    inline ll aug(int u, ll Flow)
    {
        if(u == T) return Flow;
        ll flow = 0;
        for(int &i = cur[u], v; ~i; i=edge[i].nex)
        {
            v = edge[i].to;
            if(deep[u] == deep[v] + 1)
            {
                ll tmp = aug(v, min(edge[i].flow, Flow));
                flow += tmp; Flow -= tmp; edge[i].flow -= tmp; edge[i ^ 1].flow += tmp;
                if(!Flow) return flow;
            }
        }
        if(!(--gap[deep[u]])) deep[S] = node + 1;
        ++gap[++deep[u]]; cur[u] = head[u];
        return flow;
    }
    inline ll Max_Flow()
    {
        init();
        ll ret = aug(S, INF);
        while(deep[S] <= node) ret += aug(S, INF);
        return ret;
    }
} mf;
int s, t;
inline void init()
{
    cnt = 0; s = 0; t = N + 1; mf.S = N + 2; mf.T = N + 3;
    for(int i=0; i<=N + 3; i++) { head[i] = -1; du[i] = 0; }
}
int main()
{
    scanf("%d%d", &N, &M);
    init();
    for(int i=1, u, v, w, op; i<=M; i++)
    {
        scanf("%d%d%d%d", &u, &v, &w, &op);
        if(op)
        {
            _add(u, v, 0); fl[i] = w;
            du[u] -= w; du[v] += w;
        }
        else
        {
            _add(u, v, w); fl[i] = 0;
        }
    }
    sum = 0;
    for(int i=1; i<=N; i++)
    {
        if(du[i] > 0)
        {
            _add(mf.S, i, du[i]);
            sum += du[i];
        }
        else if(du[i] < 0)
        {
            _add(i, mf.T, -du[i]);
            sum -= du[i];
        }
    }
    _add(s, 1, INF); _add(N, t, INF);
    _add(t, s, INF); mf.node = N + 4;
    if(mf.Max_Flow() * 2LL != sum) { printf("Impossible\n"); return 0; }
    ll tmp = edge[cnt - 1].flow; edge[cnt - 2].flow = edge[cnt - 1].flow = 0;
    head[mf.S] = head[mf.T] = -1;
    mf.S = t; mf.T = s; mf.node = N + 2;
    ll ans = tmp - mf.Max_Flow();
    printf("%lld\n", ans);
    for(int i=1; i<=M; i++)
    {
        printf("%lld%c", edge[i * 2 - 1].flow + fl[i], i == M ? '\n' : ' ');
    }
    return 0;
}