P4452 [国家集训队]航班安排
解题思路：
感觉题面让人有很多误解，就是说有k架飞机在0点从0号机场起飞，在t时刻返回机场，给出空载飞行的时间和花费以及m个包机请求的花费和起始时间和终止时间，想要得到净利润最大，显然是个费用流，对于每个询问拆成两个点，点之间连一条流量为1，花费为这次请求的c，由于从0号机场起飞，如果满足0+从0号机场飞到该次请求的机场a的时间<=请求任务开始的时间s，那么建一条流量为inf，费用为-从0号机场飞到该次请求的机场a的花费，同样如果满足该次请求的t+该次请求的b机场飞到0号机场的时间<=t，也建一条流量为inf，费用为-从该次请求的b号机场飞到0号机场的花费，最后对于每个询问，是否能与其他询问连边，如果一个询问的结束时间t+从该询问的b机场到另一个询问的a机场的时间<=另一个询问的开始时间s，那么也能连上一条流量为inf,花费为-空载飞行的花费的边，最后由于有k架飞机，再建个超级源点流到0号机场的点，流量为k，花费为0即可
代码：
代码写得和解题思路有点出入，主要是点位序号我为了方便把0号机场变成了1号
#include <bits/stdc++.h>
using namespace std;
/* freopen("k.in", "r", stdin);
freopen("k.out", "w", stdout); */
// clock_t c1 = clock();
// std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define ls ((x) << 1)
#define rs ((x) << 1 | 1)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 1e5 + 7;
const ll MAXM = 1e6 + 7;
const ll MOD = 1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1.0);
inline int read_int()
{
char c;
int ret = 0, sgn = 1;
do
{
c = getchar();
} while ((c < '0' || c > '9') && c != '-');
if (c == '-')
sgn = -1;
else
ret = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
ret = ret * 10 + (c - '0');
return sgn * ret;
}
inline ll read_ll()
{
char c;
ll ret = 0, sgn = 1;
do
{
c = getchar();
} while ((c < '0' || c > '9') && c != '-');
if (c == '-')
sgn = -1;
else
ret = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
ret = ret * 10 + (c - '0');
return sgn * ret;
}
struct edge
{
int to, capacity, cost, rev;
edge() {}
edge(int to, int _capacity, int _cost, int _rev) : to(to), capacity(_capacity), cost(_cost), rev(_rev) {}
};
struct Min_Cost_Max_Flow
{
int V, H[MAXN + 5], dis[MAXN + 5], PreV[MAXN + 5], PreE[MAXN + 5];
vector<edge> G[MAXN + 5];
void Init(int n)
{
V = n;
for (int i = 0; i <= V; ++i)
G[i].clear();
}
void Add_Edge(int from, int to, int cap, int cost)
{
G[from].push_back(edge(to, cap, cost, G[to].size()));
G[to].push_back(edge(from, 0, -cost, G[from].size() - 1));
}
int Min_cost_max_flow(int s, int t, int f, int &flow)
{
int res = 0;
fill(H, H + 1 + V, 0);
while (f)
{
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
fill(dis, dis + 1 + V, inf);
dis[s] = 0;
q.push(pair<int, int>(0, s));
while (!q.empty())
{
pair<int, int> now = q.top();
q.pop();
int v = now.second;
if (dis[v] < now.first)
continue;
for (int i = 0; i < G[v].size(); ++i)
{
edge &e = G[v][i];
if (e.capacity > 0 && dis[e.to] > dis[v] + e.cost + H[v] - H[e.to])
{
dis[e.to] = dis[v] + e.cost + H[v] - H[e.to];
PreV[e.to] = v;
PreE[e.to] = i;
q.push(pair<int, int>(dis[e.to], e.to));
}
}
}
if (dis[t] == inf)
break;
for (int i = 0; i <= V; ++i)
H[i] += dis[i];
int d = f;
for (int v = t; v != s; v = PreV[v])
d = min(d, G[PreV[v]][PreE[v]].capacity);
f -= d;
flow += d;
res += d * H[t];
for (int v = t; v != s; v = PreV[v])
{
edge &e = G[PreV[v]][PreE[v]];
e.capacity -= d;
G[v][e.rev].capacity += d;
}
}
return res;
}
int Max_cost_max_flow(int s, int t, int f, int &flow)
{
int res = 0;
fill(H, H + 1 + V, 0);
while (f)
{
priority_queue<pair<int, int>> q;
fill(dis, dis + 1 + V, -inf);
dis[s] = 0;
q.push(pair<int, int>(0, s));
while (!q.empty())
{
pair<int, int> now = q.top();
q.pop();
int v = now.second;
if (dis[v] > now.first)
continue;
for (int i = 0; i < G[v].size(); ++i)
{
edge &e = G[v][i];
if (e.capacity > 0 && dis[e.to] < dis[v] + e.cost + H[v] - H[e.to])
{
dis[e.to] = dis[v] + e.cost + H[v] - H[e.to];
PreV[e.to] = v;
PreE[e.to] = i;
q.push(pair<int, int>(dis[e.to], e.to));
}
}
}
if (dis[t] == -inf)
break;
for (int i = 0; i <= V; ++i)
H[i] += dis[i];
int d = f;
for (int v = t; v != s; v = PreV[v])
d = min(d, G[PreV[v]][PreE[v]].capacity);
f -= d;
flow += d;
res += d * H[t];
for (int v = t; v != s; v = PreV[v])
{
edge &e = G[PreV[v]][PreE[v]];
e.capacity -= d;
G[v][e.rev].capacity += d;
}
}
return res;
}
};
Min_Cost_Max_Flow MCMF;
PII p[305][305];
struct quest
{
int a, b, s, t, c;
} q[MAXN + 5];
int main()
{
int n = read_int(), m = read_int(), k = read_int(), t = read_int();
MCMF.Init((m << 1 | 1) + 1);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
p[i][j].first = read_int();
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
p[i][j].second = read_int();
for (int i = 1; i <= m; i++)
{
q[i].a = read_int(), q[i].b = read_int(), q[i].s = read_int(), q[i].t = read_int(), q[i].c = read_int();
q[i].a++, q[i].b++;
}
for (int i = 1; i <= m; i++)
{
MCMF.Add_Edge(i << 1, i << 1 | 1, 1, q[i].c);
if (p[1][q[i].a].first <= q[i].s)
MCMF.Add_Edge(1, i << 1, inf, -p[1][q[i].a].second);
if (p[q[i].b][1].first + q[i].t <= t)
MCMF.Add_Edge(i << 1 | 1, (m << 1 | 1) + 1, inf, -p[q[i].b][1].second);
for (int j = 1; j <= m; j++)
{
if (i == j)
continue;
if (q[i].t + p[q[i].b][q[j].a].first <= q[j].s)
MCMF.Add_Edge(i << 1 | 1, j << 1, inf, -p[q[i].b][q[j].a].second);
}
}
int st = 0, ed = (m << 1 | 1) + 1;
MCMF.Add_Edge(st, 1, k, 0);
int flow = 0;
int ans = MCMF.Max_cost_max_flow(st, ed, inf, flow);
printf("%d
", ans);
return 0;
}