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）

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("%lldn", flow);
return 0;
}