【算法练习】Luogu P2604 [ZJOI2010]网络扩容（最大流+费用流）

题意

给定一张有向图，每条边都有一个容量C和一个扩容费用W。这里扩容费用是指将容量扩大1所需的费用。求：

1. 在不扩容的情况下，1到N的最大流；
2. 将1到N的最大流增加K所需的最小扩容费用。

题解

第一问最大流。
第二问，在原来最大流参量网络基础上，考虑如何加边求出结果。因为要求在原来的基础上增加流量K，新建源点S，S连1，容量为K，费用为0，这样保证了新增的流量为K。在原来的边的基础上，增加边，容量为INF，费用为W。这样跑从S到N的最小费用最大流，费用即为结果。

代码

//
// Created by pengwill on 2018/9/26.
//
#include 
using namespace std;
typedef int valtype;
const int nmax = 50007;
const int INF = 0x3f3f3f3f;
struct ee {
    int u, v, c, w;
}add[nmax];
vector<ee> nv;
struct MCMF{
    valtype final_flow,final_cost;
    int tot,S,T;
    bool inque[nmax];
    int head[nmax], pre_edge[nmax],pre_index[nmax];
    valtype add_flow[nmax], dis[nmax];
    struct edge{int to,nxt;valtype cap,flow,cost;}e[nmax<<1];
    void init(int S, int T){
//        memset(head,-1,sizeof head);
//        tot = 0;
//        final_cost = final_flow = 0;
        this->S = S, this->T = T;
    }
    void add_edge(int u, int v, valtype cap, valtype cost){
        e[tot].to = v, e[tot].nxt = head[u], e[tot].flow = 0, e[tot].cap = cap, e[tot].cost = cost, head[u] = tot++;
        e[tot].to = u, e[tot].nxt = head[v], e[tot].flow = 0, e[tot].cap = 0, e[tot].cost = -cost, head[v] = tot++;
    }
    bool spfa(){
        for(int i = S;i<=T;++i) inque[i] = false, dis[i] = i == S?0:INF;
        queue<int> q; q.push(S), inque[S] = true, add_flow[S] = INF;
        while(!q.empty()){
            int u = q.front(); q.pop(); inque[u] = false;
            for(int i = head[u];i!=-1;i=e[i].nxt){
                int v = e[i].to;
                if(e[i].cap > e[i].flow && dis[u] + e[i].cost < dis[v]){
                    dis[v] = dis[u] + e[i].cost, pre_edge[v] = i, pre_index[v] = u;
                    add_flow[v] = min(add_flow[u],e[i].cap - e[i].flow);
                    if(!inque[v]) q.push(v),inque[v] = true;
                }
            }
        }
        return dis[T] != INF;
    }
    void mincost_mxflow() {
        while(spfa()){
            final_flow += add_flow[T];
            final_cost += add_flow[T] * dis[T];
            int now = T;
            while(now != S){
                e[pre_edge[now]].flow += add_flow[T];
                e[pre_edge[now]^1].flow -= add_flow[T];
                now = pre_index[now];
            }
        }
    }
}mcmf;
int n, m, k;
int main() {
    memset(mcmf.head, -1, sizeof(mcmf.head));
    int s = 0, t = n;
    scanf("%d %d %d", &n, &m, &k);
    int u, v, c, w;
    mcmf.S = 1, mcmf.T = n;
    for(int i = 1; i <= m; ++i) {
        scanf("%d %d %d %d", &u, &v, &c, &w);
        nv.push_back(ee{u, v, c, w});
        mcmf.add_edge(u, v, c, 0);
    }
    mcmf.mincost_mxflow();
    printf("%d ", mcmf.final_flow);
    mcmf.final_flow = mcmf.final_cost = 0;
    for(int i = 0; i < nv.size(); ++i) {
        mcmf.add_edge(nv[i].u, nv[i].v, INF, nv[i].w);
    }
    mcmf.S = 0, mcmf.T = n;
    mcmf.add_edge(0, 1, k, 0);
    mcmf.mincost_mxflow();
    printf("%d\n", mcmf.final_cost);
}