The Chinese Postman Problem【HIT-2739】【中国邮路问题/最小费用可行流】

题目链接 HIT 2739

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有上下界网络流

  每条边至少经过一次，但是不限制经过次数，这么不就是有下界网络流的做法嘛，首先，将每个点的入度和出度确定下来，然后呢，入度减出度大于0的就和源点连接，入度减出度小于0的就和汇点连接，于是就建立起了有上下界网络流模型。

  然后，判断“-1”，有入度无出度，或者有出度无入度的都是不可行的。

#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 0x3f3f3f3f
#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)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e2 + 7, maxM = 1e4 + 7;
int N, M, head[maxN], cnt, In_Out[maxN];
bool In[maxN], Ou[maxN];
struct Eddge
{
    int nex, u, v; int flow, cost;
    Eddge(int a=-1, int b=0, int c=0, int d=0, int f=0):nex(a), u(b), v(c), flow(d), cost(f) {}
} edge[maxM];
inline void addEddge(int u, int v, int f, int w)
{
    edge[cnt] = Eddge(head[u], u, v, f, w);
    head[u] = cnt++;
}
inline void _add(int u, int v, int f, int w) { addEddge(u, v, f, w); addEddge(v, u, 0, -w); }
struct MaxFlow_MinCost
{
    int pre[maxN], S, T; int Flow[maxN], dist[maxN];
    queue<int> Q;
    bool inque[maxN];
    inline bool spfa()
    {
        for(int i=0; i<=T; i++) { pre[i] = -1; dist[i] = INF; inque[i] = false; }
        while(!Q.empty()) Q.pop();
        Q.push(S); dist[S] = 0; inque[S] = true; Flow[S] = INF;
        while(!Q.empty())
        {
            int u = Q.front(); Q.pop(); inque[u] = false;
            int f, w;
            for(int i=head[u], v; ~i; i=edge[i].nex)
            {
                v = edge[i].v; f = edge[i].flow; w = edge[i].cost;
                if(f && dist[v] > dist[u] + w)
                {
                    dist[v] = dist[u] + w;
                    Flow[v] = min(Flow[u], f);
                    pre[v] = i;
                    if(!inque[v])
                    {
                        inque[v] = true;
                        Q.push(v);
                    }
                }
            }
        }
        return ~pre[T];
    }
    inline int EK()
    {
        int sum_Cost = 0;
        while(spfa())
        {
            int now = T, las = pre[now];
            while(now ^ S)
            {
                edge[las].flow -= Flow[T];
                edge[las ^ 1].flow += Flow[T];
                now = edge[las].u;
                las = pre[now];
            }
            sum_Cost += dist[T] * Flow[T];
        }
        return sum_Cost;
    }
} MF;
void init()
{
    cnt = 0; MF.S = 0; MF.T = N + 1;
    for(int i=0; i<=MF.T; i++) head[i] = -1;
    for(int i=0; i<N; i++) { In_Out[i] = 0; In[i] = Ou[i] = false; }
}
int main()
{
    int Cas; scanf("%d", &Cas);
    while(Cas--)
    {
        scanf("%d%d", &N, &M);
        init();
        int ans = 0;
        for(int i=1, u, v, w; i<=M; i++)
        {
            scanf("%d%d%d", &u, &v, &w);
            ans += w;
            _add(u, v, INF, w);
            Ou[u] = In[v] = true;
            In_Out[u] --; In_Out[v] ++;
        }
        bool ok = true;
        for(int i=0; i<N; i++)
        {
            if((!In[i] && Ou[i])|| (!Ou[i] && In[i])) { ok = false; break; }
        }
        if(!ok)
        {
            printf("-1\n");
            continue;
        }
        for(int i=0; i<N; i++)
        {
            if(In_Out[i] > 0)
            {
                _add(MF.S, i, In_Out[i], 0);
            }
            else if(In_Out[i] < 0)
            {
                _add(i, MF.T, -In_Out[i], 0);
            }
        }
        ans += MF.EK();
        printf("%d\n", ans);
    }
    return 0;
}