CCPC 2018 秦皇岛 A题 Build（网络流、限制花费的最小费用最大流）

CCPC 2018 秦皇岛 A题 Build（网络流、限制花费的最小费用最大流）

题意：给一个无向图，给出每条边的容量和单位花费，求从1到n费用不超过f的最大流

如果当次spfa跑出来的流量已经不能全部买到了，那就切一部分买到的。

#include <bits/stdc++.h>

using namespace std;

const int maxn = 2100;
const int maxm = 2100000;
const int inf = 0x3f3f3f3f3f3f3f3f;
typedef long long ll;
struct Edge {
    int to, nex;
    ll cap, flow, cost;
} edge[maxm];


int head[maxn], tot;
int pre[maxn];
ll dis[maxn];
bool vis[maxn];
int N;

void init(int n) {
    N = n;
    tot = 0;
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v, int cap, int cost) {
    edge[tot].to = v;
    edge[tot].cap = cap;
    edge[tot].cost = cost;
    edge[tot].flow = 0;
    edge[tot].nex = head[u];
    head[u] = tot++;
    edge[tot].to = u;
    edge[tot].cap = 0;
    edge[tot].cost = -cost;
    edge[tot].flow = 0;
    edge[tot].nex = head[v];
    head[v] = tot++;
}


bool spfa(int s, int t) {
    queue<int> q;
    for (int i = 0; i <= N; i++) {
        dis[i] = inf;
        vis[i] = false;
        pre[i] = -1;
    }

    dis[s] = 0;
    vis[s] = true;

    q.push(s);
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        vis[u] = false;

        for (int i = head[u]; i != -1; i = edge[i].nex) {
            int v = edge[i].to;
            if (edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost) {
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if (!vis[v]) {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    return pre[t] != -1;
}

int n, m;
ll f;

ll mcmf(int s, int t, ll &cost, ll flow) {
    cost = 0;
    while (spfa(s, t)) {
        ll Min = inf;
        ll cost1 = 0;
        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]) {
            if (Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]) {
            edge[i].flow += Min;
            edge[i ^ 1].flow -= Min;
            cost1 += edge[i].cost;
        }
        ll tmp = (f - cost) / cost1;
        if (Min > tmp)
            return flow + tmp;
        cost += cost1 * Min;
        flow += Min;
    }
    return flow;
}


int mx[maxm], costi[maxm];
int from[maxm], to[maxm];


int main() {
    scanf("%d %d %lld", &n, &m, &f);
    init(2 * n + 1);
    addedge(0, 1, inf, 0);
    for (int i = 1; i <= n; i++) {
        addedge(i, i + n, inf, 0);
    }
    for (int i = 1; i <= m; i++) {
        scanf("%d %d %d %d", &from[i], &to[i], &mx[i], &costi[i]);
        addedge(from[i] + n, to[i], mx[i], costi[i]);
        addedge(to[i] + n, from[i], mx[i], costi[i]);
    }
    ll cost, flow;
    flow = mcmf(0, 2 * n, cost, 0);
    printf("%lld
", flow);
    return 0;
}