80人环游世界【有上下界最小费用可行流】

题目链接 P4553 80人环游世界

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  我们把城市拆点，保证每个点都是被经过了Vi次。然后保证只有最多M个人去参观，所以我们要做到限流，我将限流放在源点上，建立一个和建立边从到流量为M，就是保证了只有M个人。

  剩下的就是有上下界网络流的基础构图了，这道题是有源汇的，所以别忘了要把原图中的源汇店要建立从汇点到源点的流量为INF，费用为0的边。然后把入流大于出流的与超级源点相连接，将入流小于出流的和超级汇点相连接，保证了下限，主要是拆点的两个点。

#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 0x3f3f3f
#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 = 2e2 + 7, maxM = 2e4 + 7;
int N, M, V[maxN], head[maxN], cnt, ps, s, t;
struct Eddge
{
    int nex, u, v, flow, cost;
    Eddge(int a=-1, int _u=0, int _v = 0, int c=0, int d=0):nex(a), u(_u), v(_v), flow(c), cost(d) {}
}edge[maxM];
inline void addEddge(int u, int v, int flow, int cost)
{
    edge[cnt] = Eddge(head[u], u, v, flow, cost);
    head[u] = cnt++;
}
inline void _add(int u, int v, int flow, int cost) { addEddge(u, v, flow, cost); addEddge(v, u, 0, -cost); }
struct MaxFlow_MinCost
{
    int pre[maxN], S, T; ll 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;
            ll 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 ll EK()
    {
        ll 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;
inline void init()
{
    cnt = 0; s = 2 * N + 1; ps = 0; t = s + 1; MF.S = t + 1; MF.T = MF.S + 1;
    for(int i=0; i<=MF.T; i++) head[i] = -1;
}
int main()
{
    scanf("%d%d", &N, &M);
    init();
    _add(s, ps, M, 0);
    for(int i=1; i<=N; i++)
    {
        scanf("%d", &V[i]); //Vi~Vi(Vi - Vi == 0)
        _add(MF.S, N + i, V[i], 0);
        _add(i, MF.T, V[i], 0);
        _add(ps, i, M, 0);
        _add(N + i, t, M, 0);
    }
    for(int i=1, Wij; i<N; i++)
    {
        for(int j = i + 1; j<=N; j++)
        {
            scanf("%d", &Wij);
            if(~Wij) _add(N + i, j, INF, Wij);
        }
    }
    _add(t, s, INF, 0);
    printf("%lld\n", MF.EK());
    return 0;
}