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题意
思路
本题做法,给每个月建立两个点 i 和 i',在建立源点s和汇点t。
①源点s向每个点 i 连一条边,流量为第i个月的产量,花费为生产成本
②每个点 i' 到汇点建一条边,流量为第i个月的销量,花费为(负的!)单位售价
③每个点 i ,向 i' ,(i'+1),(i'+2),(i'+3)...连一条边,流量为INF,花费为储存月数的花费
当s-t的最短路长度大于0时,就停止增广,因为再增广就会使最大利润减小了。
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
const int INF = 0x3f3f3f3f;
struct Edge
{
int from, to, cap, flow, cost; //起点,终点,容量,流量,花费
Edge(int u, int v, int c, int f, int w):from(u), to(v), cap(c), flow(f), cost(w) {}
};
struct MCMF
{
int n, m; //结点数,边数(包括反向弧),源点s,汇点t
vector<Edge> edges; //边表。edges[e]和edges[e^1]互为反向弧
vector<int> G[MAXN]; //邻接表,G[i][j]表示结点i的第j条边在edges数组中的序号
bool inq[MAXN]; //是否在队列中
int d[MAXN]; //Bellman-Ford
int p[MAXN]; //上一条弧
int a[MAXN]; //可改进量
void init(int n)
{
this->n = n;
edges.clear();
for (int i = 0; i <= n; i++) G[i].clear();
}
void AddEdge(int from, int to, int cap, int cost)
{
edges.push_back(Edge(from, to, cap, 0, cost));
edges.push_back(Edge(to, from, 0, 0, -cost));
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool BellmanFord(int s, int t, int &flow, long long& cost)//SPFA
{
for (int i = 0; i <= n; i++) d[i] = INF;
memset(inq, 0, sizeof(inq));
d[s] = 0; inq[s] = true; p[s] = 0; a[s] = INF;
queue<int> Q;
Q.push(s);
while (!Q.empty())
{
int u = Q.front(); Q.pop();
inq[u] = 0;
for (int i = 0; i < G[u].size(); i++)
{
Edge& e = edges[G[u][i]];
if (e.cap > e.flow && d[e.to] > d[u] + e.cost)
{
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if (!inq[e.to]) { Q.push(e.to); inq[e.to] = true; }
}
}
}
if (d[t] > 0) return false;
if (d[t] == INF) return false;
flow += a[t];
cost += (long long)d[t] * (long long)a[t];
for (int u = t; u != s; u = edges[p[u]].from)
{
edges[p[u]].flow += a[t];
edges[p[u]^1].flow -= a[t];
}
return true;
}
int MinCostMaxFlow(int s, int t, long long& cost)
{
int flow = 0; cost = 0;
while (BellmanFord(s, t, flow, cost));
return flow;
}
}solve;
int main()
{
int T, CASE = 1; scanf("%d", &T);
while(T--)
{
int M, I; scanf("%d%d", &M, &I);
int S = 0, T = 2*M+1;
solve.init(T);
for (int i = 1; i <= M; i++)
{
int a, b, c, d, e;
scanf("%d%d%d%d%d", &a, &b, &c, &d, &e);
solve.AddEdge(S, i, b, a);
for (int j = 0; j <= e; j++) if (i+j <= M) solve.AddEdge(i, i+j+M, INF, I*j);
solve.AddEdge(i+M, T, d, -c);
}
long long cost = 0;
int Max_Flow = solve.MinCostMaxFlow(S, T, cost);
printf("Case %d: %lld\n", CASE++, -cost);
}
return 0;
}
/*
1
2 2
2 10 3 20 2
10 100 7 5 2
*/